De nicijaszunar/m1_materijali...sinx ne postoji. Primjer 1(f) sin 1 x 1 1 x y lim x!0 + sin 1 x i...

�De�nicija�

Neka je f : D ⊆ R→ R. Za c ∈ R ∪ {±∞}, ako vrijedi

x → c︸︷︷︸�x se pribliºava c,

x 6= c�

⇒ f (x)→ L︸︷︷︸�f (x) se pribliºava L�

za neki L ∈ R ∪ {±∞}, pi²emo

limx→c

f (x) = L.

Ako je L ∈ R, kaºemo da je L limes funkcije f u to£ki c .

Primjer 1(a)

limx→4

f (x) =73

Primjer 1(b)

y Γex

limx→−∞

ex = 0, limx→+∞

ex = +∞, limx→0

ex = 1(= e0).

Primjer 1(c)

limx→−2

f (x) = +∞.

Primjer 1(d)

limx→3

f (x) ne postoji, ali limx→3−

f (x) = −∞, a limx→3+

f (x) = 1.

�De�nicija�

Neka je f : D ⊆ R→ R. Za c ∈ R, ako vrijedi

x → c+︸︷︷︸�x se pribliºava c,

x > c�

limx→c+

f (x) = L.

Ako je L ∈ R, kaºemo da je L limes zdesna funkcije f u to£ki c .

�De�nicija�

Neka je f : D ⊆ R→ R. Za c ∈ R, ako vrijedi

x → c−︸︷︷︸�x se pribliºava c,

x<c�

limx→c−

f (x) = L.

Ako je L ∈ R, kaºemo da je L limes slijeva funkcije f u to£ki c .

Primjer 1(e)

limx→+∞

sin x ne postoji.

Primjer 1(f)

Γsin 1

limx→0+

xi lim

x→0−sin

xne postoje.

Primjer 1(g)

limx→−2−

ln x i limx→−2+

ln x ne postoje.

Limesi i osnovne aritmeti£ke operacije

Neka su a ∈ R i c ∈ R ∪ {±∞}. Svaka od sljede¢ih jednakosti vrijedi kad god je njezina desna

strana de�nirana:

(i) limx→c

(f (x) + g(x)) = limx→c

f (x) + limx→c

(ii) limx→c

f (x) · g(x) = limx→c

f (x) · limx→c

(iii) limx→c

a f (x) = a · limx→c

(iv) limx→c

limx→c f (x)

limx→c g(x).

Analogne tvrdnje vrijede i za limx→c+ i limx→c−.

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

(limx→+∞ 3 + limx→+∞

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

)limx→+∞ 3 + limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

= limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(a)

limx→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1=

(+∞+∞

)= lim

x→+∞

(3x + 5)(2x − 3)(4x − 7)

3x3 + x − 1·

= limx→+∞

3x+5x · 2x−3x · 4x−7x

3 + 1x2− 1

= limx→+∞

(3 + 5

) (2− 3

) (4− 7

)3 + 1

x2− 1

) (limx→+∞ 2− limx→+∞

) (limx→+∞ 4− limx→+∞

1x2− limx→+∞

=(3 + 0)(2− 0)(4− 0)

3 + 0− 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Pitanje. Smije li limx→+∞ u¢i pod 3√

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

= limx→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Primjer 2(b)

limx→+∞

x3√x3 + 10

(+∞+∞

)= lim

x→+∞

x3√x3 + 10

·1x1x

= limx→+∞

13√x3+103√x3

= limx→+∞

√1 + 10

√1 + limx→+∞

3√1 + 0

Smije li lim u¢i pod 3

Vrijedi

limx→c

g(f (x)) = g(

limx→c

f (x))

kad god su zadovoljeni sljede¢i uvjeti:

g je elementarna funkcija: polinom, racionalna funkcija, n√· , trigonometrijska, arkus,

eksponencijalna, logaritamska funkcija ili | · |.Desna strana formule (1) je de�nirana.

g(f (x)) je de�nirano za x iz nekog probu²enog intervala oko c , tj. intervala oblikac

, ako je c ∈ R,

c, ako je c = +∞,

, ako je c = −∞.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

√−x2 ne postoji.

Naime, D√−x2 = {0} pa se Γ√−x2 sastoji samo od jedne to£ke:

Γ√−x2

(b) limx→+∞

arctg1

= arctg limx→+∞

x= arctg 0 = 0.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

= arctg limx→+∞

x= arctg 0 = 0.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

= arctg limx→+∞

x= arctg 0 = 0.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

= arctg limx→+∞

x= arctg 0 = 0.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

x= arctg lim

x→+∞

= arctg 0 = 0.

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

x= arctg lim

x→+∞

x= arctg 0

Primjer 3

(a) Oprez!

limx→0

√−x2 6=

√limx→0

(−x2) =√0 = 0.

Zapravo, limx→0

Γ√−x2

(b) limx→+∞

arctg1

x= arctg lim

x→+∞

x= arctg 0 = 0.

Odre�eni i neodre�eni oblici

Primjer. limx→1−

ln(1− x)

(x − 1)2

(−∞0+

)= −∞.

−∞0+

je tzv. odre�eni oblik � moºe se izra£unati (i iznosi −∞). S druge strane,

∞∞, (+∞) + (−∞), ∞ · 0

su tzv. neodre�eni oblici � ne mogu se jednozna£no odrediti.

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

→0+︷︸︸︷(1− x)

(x − 1)2

(−∞0+

)= −∞.

−∞0+

∞∞, (+∞) + (−∞), ∞ · 0

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

→0+︷︸︸︷(1− x)

(x − 1)2=

(−∞0+

= −∞.

−∞0+

∞∞, (+∞) + (−∞), ∞ · 0

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

→0+︷︸︸︷(1− x)

(x − 1)2=

(−∞0+

)= −∞.

−∞0+

∞∞, (+∞) + (−∞), ∞ · 0

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

→0+︷︸︸︷(1− x)

(x − 1)2=

(−∞0+

)= −∞.

−∞0+

je tzv. odre�eni oblik � moºe se izra£unati (i iznosi −∞).

S druge strane,

∞∞, (+∞) + (−∞), ∞ · 0

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

→0+︷︸︸︷(1− x)

(x − 1)2=

(−∞0+

)= −∞.

−∞0+

∞∞, (+∞) + (−∞), ∞ · 0

Primjer. limx→0

)= lim

limx→0

)= lim

x→0x = 0

limx→0

)= lim

x2= +∞

limx→0

)= lim

xne postoji.

Neki odre�eni oblici

Neka je a ∈ R. Vrijedi:

∞= 0

{+∞, ako je a > 0

−∞, ako je a < 0,

{−∞, ako je a > 0

+∞, ako je a < 0.

+∞0+

= +∞, +∞0−

= −∞, −∞0+

= −∞, −∞0−

= +∞,

(+∞) + (+∞) = +∞, (−∞) + (−∞) = −∞

(+∞) · (+∞) = +∞, (−∞) · (−∞) = +∞, (+∞) · (−∞) = −∞.

Zadatak 36(a)

Izra£unajte sljede¢i limes (ako postoji): limx→+∞

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

= limx→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(a)

(x + 1)2

x2 + 1.

Rje²enje. Imamo

limx→+∞

(x + 1)2

x2 + 1=

(+∞+∞

)= lim

x→+∞

(x + 1)2

x2 + 1·

= limx→+∞

)21 + 1

= limx→+∞

(1 + 1

)21 + 1

=(1 + 0)2

Zadatak 36(b)

Izra£unajte sljede¢i limes (ako postoji):

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1=

(+∞+∞

= limx→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1=

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1=

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1=

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(b)

limx→+∞

x2 − 1.

Rje²enje. Imamo

limx→+∞

x2 − 1=

(+∞+∞

)= lim

x→+∞

x2 − 1·

= limx→+∞

1− 1x2

1− 0

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7=

(+∞+∞

= limx→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7=

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7=

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7=

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

= +∞.

Zadatak 36(c)

limx→+∞

x2 − 5x + 1

3x + 7.

Rje²enje. Imamo

limx→+∞

x2 − 5x + 1

3x + 7=

(+∞+∞

)= lim

x→+∞

x2 − 5x + 1

3x + 7·

= limx→+∞

x − 5 + 1x

3 + 7x

(+∞3

)= +∞.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

pri £emu smo u tre¢oj jednakosti iskoristili da vrijedi√x4 + 1

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

= limx→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(d)

limx→−∞

2x2 − 3x − 4√x4 + 1

Rje²enje. Imamolim

x→−∞

2x2 − 3x − 4√x4 + 1

(+∞+∞

)= lim

x→−∞

2x2 − 3x − 4√x4 + 1

= limx→−∞

2− 3x −

4x2√

1 + 1x4

=2− 0− 0√

√x4 + 1√x4

√x4 + 1

√1 +

x4, x ∈ R \ {0}.

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

pri £emu smo u tre¢oj jednakosti iskoristili da vrijedi√x +

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

= limx→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 36(e)

√x√

x +√x +√x.

Rje²enje. Imamolim

x→+∞

√x√

x +√

x +√x

(+∞+∞

)= lim

x→+∞

√x√

x +√x +√x·

= limx→+∞

1√1 +

√1x + 1

1 +√0 + 0

√x +√x

√x +

√x +√x

√1 +

√x +√x√

√1 +

√x +√x

√√√√1 +

, x ∈ 〈0,+∞〉 .

Zadatak 37(a)

Izra£unajte sljede¢i limes (ako postoji): limx→2

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2

)= lim

(x + 2)

(x − 1)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2

)= lim

(x + 2)

(x − 1)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

= limx→2

(x + 2)

(x − 1)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

)= lim

(x − 2)(x + 2)

(x − 1)(x − 2)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

)= lim

��(x − 2)(x + 2)

(x − 1)��(x − 2)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

)= lim

��(x − 2)(x + 2)

(x − 1)��(x − 2)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

)= lim

��(x − 2)(x + 2)

(x − 1)��(x − 2)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(a)

x2 − 4

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 4

x2 − 3x + 2=

)= lim

��(x − 2)(x + 2)

(x − 1)��(x − 2)

= limx→2

x − 1

=2 + 2

2− 1

Zadatak 37(b)

x2 − 4

x − 1.

Rje²enje. Imamo

limx→2

x2 − 4

x − 1

=22 − 4

2− 1= 0.

Zadatak 37(b)

x2 − 4

x − 1.

Rje²enje. Imamo

limx→2

x2 − 4

x − 1

=22 − 4

2− 1= 0.

Zadatak 37(b)

x2 − 4

x − 1.

Rje²enje. Imamo

limx→2

x2 − 4

x − 1=

22 − 4

2− 1

Zadatak 37(b)

x2 − 4

x − 1.

Rje²enje. Imamo

limx→2

x2 − 4

x − 1=

22 − 4

2− 1= 0.

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 3x + 2

)= lim

(x − 1)(x − 2)

1 · 0

)= ±∞,

dakle traºeni limes ne postoji.

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 3x + 2

)= lim

(x − 1)(x − 2)

1 · 0

)= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 3x + 2=

= limx→2

(x − 1)(x − 2)

1 · 0

)= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 3x + 2=

)= lim

(x − 1)(x − 2)

1 · 0

)= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2

x2 − 3x + 2=

)= lim

(x − 1)(x − 2)

1 · 0

= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2±

x2 − 3x + 2=

)= lim

x→2±

(x − 1)(x − 2)

1 · 0±

= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2±

x2 − 3x + 2=

)= lim

x→2±

(x − 1)(x − 2)

1 · 0±

)= ±∞,

Zadatak 37(c)

x2 − 3x + 2.

Rje²enje. Imamo

limx→2±

x2 − 3x + 2=

)= lim

x→2±

(x − 1)(x − 2)

1 · 0±

)= ±∞,

Zadatak 37(d)

Izra£unajte sljede¢i limes (ako postoji): limx→−1

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

)= lim

x→−1

(x2 − x + 1

)= lim

x→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

pri £emu smo u drugoj jednakosti primijenili formulu

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

= limx→−1

(x2 − x + 1

)= lim

x→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

= limx→−1

(x2 − x + 1

)= lim

x→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

)= lim

x→−1

(x + 1)(x2 − x + 1

)x + 1

= limx→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

)= lim

x→−1��(x + 1)

(x2 − x + 1

)��x + 1

= limx→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

)= lim

x→−1��(x + 1)

(x2 − x + 1

)��x + 1

= limx→−1

(x2 − x + 1

= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

)= lim

x→−1��(x + 1)

(x2 − x + 1

)��x + 1

= limx→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(d)

x3 + 1

x + 1.

Rje²enje. Imamo

limx→−1

x3 + 1

x + 1=

)= lim

x→−1��(x + 1)

(x2 − x + 1

)��x + 1

= limx→−1

(x2 − x + 1

)= (−1)2 − (−1) + 1

a3 + b3 = (a + b)(a2 − ab + b2

), a, b ∈ R.

Zadatak 37(e)

x − 2√x −√2.

Rje²enje. 1. na£in. Imamo

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

= limx→2

(√x)2 − (√2)2√x −√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

(√x −√2) (√

x +√2)

√x −√2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

��(√x −√2) (√

x +√2)

��√x −√2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

��(√x −√2) (√

x +√2)

��√x −√2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

��(√x −√2) (√

x +√2)

��√x −√2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

(√x)2 − (√2)2√x −√2

= limx→2

��(√x −√2) (√

x +√2)

��√x −√2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

Rje²enje. 2. na£in. Racionalizacija: imamo

limx→2

x − 2√x −√2

= limx→2

x − 2√x −√2·√x +√2

√x +√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

(√x +√2)

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

(x − 2)(√

x +√2)

x − 2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

��(x − 2)(√

x +√2)

��x − 2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

��(x − 2)(√

x +√2)

��x − 2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

��(x − 2)(√

x +√2)

��x − 2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(e)

x − 2√x −√2.

limx→2

x − 2√x −√2

)= lim

x − 2√x −√2·√x +√2

√x +√2

= limx→2

��(x − 2)(√

x +√2)

��x − 2

= limx→2

(√x +√2)

=√2 +√2

= 2√2.

Zadatak 37(f)

Za a ∈ 〈0,+∞〉, izra£unajte sljede¢i limes (ako postoji): limx→a

√x −√a

x − a.

limx→a

√x −√a

x − a=

= limx→a

√x −√a(√

x)2 − (√a)2

= limx→a

(√x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

= limx→a

√x −√a(√

x)2 − (√a)2

= limx→a

(√x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

(√x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

√x −√a(√

x −√a) (√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

��√x −√a

��(√x −√a) (√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

��√x −√a

��(√x −√a) (√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

��√x −√a

��(√x −√a) (√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a(√

x)2 − (√a)2

= limx→a

��√x −√a

��(√x −√a) (√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

= limx→a

√x −√a

x − a·√x +√a√

x +√a

= limx→a

(√x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

(√x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

x − a

(x − a)(√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

��x − a

��(x − a)(√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

��x − a

��(x − a)(√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

��x − a

��(x − a)(√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(f)

√x −√a

x − a.

limx→a

√x −√a

x − a=

)= lim

√x −√a

x − a·√x +√a√

x +√a

= limx→a

��x − a

��(x − a)(√

x +√a)

= limx→a

1√x +√a

a +√a

2√a.

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

= limx→0

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

[Supstitucija: t = 6

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

(t2 + t + 1

)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

= limx→0

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

(t2 + t + 1

)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

(t2 + t + 1

)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

= limt→1

t3 − 1

t2 − 1

= limt→1

(t2 + t + 1

)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

(t2 + t + 1

)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

(t − 1)(t2 + t + 1

)(t − 1)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

��(t − 1)(t2 + t + 1

)��(t − 1)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

��(t − 1)(t2 + t + 1

)��(t − 1)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

��(t − 1)(t2 + t + 1

)��(t − 1)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

(6√1 + x

)3 − 1(6√1 + x

)2 − 1

√1 + x

x → 0 ⇒ t → 1

]= lim

t3 − 1

t2 − 1

= limt→1

��(t − 1)(t2 + t + 1

)��(t − 1)(t + 1)

= limt→1

t2 + t + 1

=12 + 1 + 1

1 + 1=

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

Rje²enje. 2. na£in.

limx→0

√1 + x − 1

3√1 + x − 1

= limx→0

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0·(

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

= limx→0

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0·(

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0·(

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0

(√1 + x

)2 − 1(3√1 + x

)3 − 1·(

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0

(√1 + x

)2 − 1(3√1 + x

)3 − 1︸︷︷︸= x

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0

(√1 + x

)2 − 1(3√1 + x

)3 − 1︸︷︷︸= x

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0

(√1 + x

)2 − 1(3√1 + x

)3 − 1︸︷︷︸= x

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Zadatak 37(g)

√1 + x − 1

3√1 + x − 1

limx→0

√1 + x − 1

3√1 + x − 1

)= lim

√1 + x − 1

3√1 + x − 1

·√1 + x + 1√1 + x + 1

3√1 + x

)2+ 3√1 + x + 1(

3√1 + x

)2+ 3√1 + x + 1

= limx→0

(√1 + x

)2 − 1(3√1 + x

)3 − 1︸︷︷︸= x

3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

= limx→0

(3√1 + x

)2+ 3√1 + x + 1

√1 + x + 1

(3√1 + 0

)2+ 3√1 + 0 + 1

√1 + 0 + 1

Vaºan limes 1

limx→0

Geometrijsko zna£enje:

sin x x

AlimB→O

|AB|∣∣∣>OB∣∣∣ = 1.

Vaºan limes 1

limx→0

Geometrijsko zna£enje:

sin x x

AlimB→O

|AB|∣∣∣>OB∣∣∣ = 1.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

= limx→0

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8= 8.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

= limx→0

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8= 8.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

)= lim

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8= 8.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

)= lim

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

= limt→0

= 1 · 8= 8.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

)= lim

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8= 8.

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

)= lim

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8

Zadatak 38(a)

sin(8x)

Rje²enje. Imamo

limx→0

sin(8x)

)= lim

sin(8x)

8x· 8

[t = 8x

x → 0 ⇒ t → 0

]= lim

= 1 · 8= 8.

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

= limx→0

sin(5x)5x · 5

sin(3x)3x · 3

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

= limx→0

sin(5x)5x · 5

sin(3x)3x · 3

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

)= lim

sin(5x)5x · 5x

sin(3x)3x · 3x

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

)= lim

sin(5x)5x · 5�x

sin(3x)3x · 3�x

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

)= lim

sin(5x)5x · 5�x

sin(3x)3x · 3�x

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

)= lim

sin(5x)5x · 5�x

sin(3x)3x · 3�x

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(b)

sin(5x)

sin(3x).

Rje²enje. Imamo

limx→0

sin(5x)

sin(3x)=

)= lim

sin(5x)5x · 5�x

sin(3x)3x · 3�x

= limx→0

sin(5x)5x · 5

sin 3x3x · 3

=1 · 51 · 3

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞

·sin π

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞

·sin π

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞

x ·sin π

· πx

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞�

x ·sin π

· π�x

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞�

x ·sin π

· π�x

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞�

x ·sin π

· π�x

= limx→+∞

sin πx

= 1 · π

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

= limx→+∞�

x ·sin π

· π�x

= limx→+∞

sin πx

= 1 · π= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

= limt→0

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1

Zadatak 38(c)

x · sinπ

limx→+∞

x · sinπ

x= ((+∞) · 0)

[t = π

x ; x = πt

x → +∞ ⇒ t → 0

]= lim

t· sin t

= limt→0

π · sin t

= π · 1= π.

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

[t = arcsin x ; x = sin t

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(d)

arcsin x

Rje²enje. Imamo

limx→0

arcsin x

x → 0 ⇒ t → 0

]= lim

= limt→0

1sin tt

Zadatak 38(e)

Rje²enje. Imamo

limx→0

= limx→0

cos x· 1x

= limx→0

= 1 · 1

cos 0= 1.

Zadatak 38(e)

Rje²enje. Imamo

limx→0

= limx→0

cos x· 1x

= limx→0

= 1 · 1

cos 0= 1.

Zadatak 38(e)

Rje²enje. Imamo

limx→0

)= lim

cos x· 1x

= limx→0

= 1 · 1

cos 0= 1.

Zadatak 38(e)

Rje²enje. Imamo

limx→0

)= lim

cos x· 1x

= limx→0

= 1 · 1

cos 0= 1.

Zadatak 38(e)

Rje²enje. Imamo

limx→0

)= lim

cos x· 1x

= limx→0

= 1 · 1

Zadatak 38(e)

Rje²enje. Imamo

limx→0

)= lim

cos x· 1x

= limx→0

= 1 · 1

cos 0= 1.

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

= limx→0

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

pri £emu smo u drugoj jednakosti iskoristili da vrijedi

cos(2t) = 1− 2 sin2 t, t ∈ R.

; Vaºan limes:limx→0

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

= limx→0

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

= limx→0

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

= limt→0

2sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

Zadatak 38(f)

1− cos x

Rje²enje. Imamo

limx→0

1− cos x

)= lim

1−(1− 2 sin2 x

= limx→0

2sin2 x

[t = x

2; x = 2t

x → 0 ⇒ t → 0

]= lim

t→02

sin2 t

= limt→0

(sin t

2· 12 =

cos(2t) = 1− 2 sin2 t, t ∈ R.

1− cos x

limx→±∞ ax

Sjetimo se:

Γax 0 < a < 1

⇒ limx→−∞

{0, ako je a > 1,

+∞, ako je 0 < a < 1,lim

x→+∞ax =

{+∞, ako je a > 1,

0, ako je 0 < a < 1.

limx→±∞ ax

Sjetimo se:

Γax 0 < a < 1

⇒ limx→−∞

{0, ako je a > 1,

+∞, ako je 0 < a < 1,lim

x→+∞ax =

{+∞, ako je a > 1,

0, ako je 0 < a < 1.

limx→c ϕ(x)ψ(x)

Kako se, za c ∈ R ∪ {±∞} i ϕ(x) > 0, ra£una limes

limx→c

ϕ(x)ψ(x)?

Vrijedi

limx→c

ϕ(x)ψ(x) =(

limx→c

ϕ(x))limx→c ψ(x)

kad god je desna strana de�nirana, tj. odre�eni oblik, primjerice:

AB sa A ∈ 〈0,+∞〉 i B ∈ R

A+∞ =

{+∞, ako je A > 1,

0, ako je 0 < A < 1

A−∞ =

{0, ako je A > 1,

+∞, ako je 0 < A < 1.

Napomena. 00, (+∞)0 i 1±∞ su neodre�eni oblici.

limx→c ϕ(x)ψ(x)

limx→c

ϕ(x)ψ(x)?

Vrijedi

limx→c

ϕ(x)ψ(x) =(

limx→c

AB sa A ∈ 〈0,+∞〉 i B ∈ R

A+∞ =

{+∞, ako je A > 1,

0, ako je 0 < A < 1

A−∞ =

{0, ako je A > 1,

+∞, ako je 0 < A < 1.

limx→c ϕ(x)ψ(x)

limx→c

ϕ(x)ψ(x)?

Vrijedi

limx→c

ϕ(x)ψ(x) =(

limx→c

AB sa A ∈ 〈0,+∞〉 i B ∈ R

A+∞ =

{+∞, ako je A > 1,

0, ako je 0 < A < 1

A−∞ =

{0, ako je A > 1,

+∞, ako je 0 < A < 1.

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

= limx→0

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

= limx→0

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2

= 1 · 2 = 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x)

= 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x) = 1 + 0

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x) = 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x) = 1 + 0 = 1

limx→0

(sin(2x)

Zadatak 39(a)

(sin(2x)

Rje²enje. Imamo

limx→0

sin(2x)

x= lim

sin(2x)

2x· 2 = 1 · 2 = 2

limx→0

(1 + x) = 1 + 0 = 1

limx→0

(sin(2x)

= 21 = 2.

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

= limx→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

= limx→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

= +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

x2 = +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

x2 = +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Zadatak 39(b)

(x + 1

2x + 1

Rje²enje. Imamo

limx→+∞

2x + 1=

(+∞+∞

)= lim

x→+∞

2x + 1·

= limx→+∞

1 + 1x

2 + 1x

=1 + 0

2 + 0=

limx→+∞

x2 = +∞

limx→+∞

(x + 1

2x + 1

)+∞)

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

=(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

=(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

x =(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

x =(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

= limt→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

x =(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

x =(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Jako vaºni limesi

Vrijedi

limx→±∞

Primjer. Imamo

limx→0±

(1 + x)1

x =(1±∞

[t = 1

x ; x = 1t

x → 0± ⇒ t → ±∞

]= lim

t→±∞

limx→0

(1 + x)1

x = e.

Primjer

Za svaki a ∈ R \ {0} imamo

limx→±∞

=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

= limt→0

(1 + t)at

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Primjer

limx→±∞

)x=(1±∞

[t = a

x ; x = at

x → ±∞ ⇒ t → 0

]= lim

t→0(1 + t)

= limt→0

((1 + t)

)a= ea,

limx→±∞

)x= ea.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

= limt→0

(1 + t)−2

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

= limt→0

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

[t = − 2

x+1; x = −2

t − 1

x → +∞ ; t → 0

]= lim

t→0(1 + t)−

t−1 = lim

((1 + t)

)−2· (1 + t)−1

= e−2 (1 + 0)−1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

= limt→0

(1 + t)1

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

= limx→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

)= lim

x→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

)= lim

x→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

)= lim

x→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

)= lim

x→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x − 1

x + 1− 1

= limx→+∞

(1− 2

= limx→+∞

((1− 2

)− x+12

)− 2xx+1

Kako jelim

x→+∞

(1− 2

)− x+12

[t = − 2

x → +∞ ⇒ t → 0

]= lim

t→0(1 + t)

x→+∞

(− 2x

)= lim

x→+∞

−2xx + 1

·1x1x

= limx→+∞

−21 + 1

=−21 + 0

= −2,

slijedi da jelim

x→+∞

(x − 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

e= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

= limx→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

e= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

e= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

=e−1

e= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

= e−2.

Zadatak 40(a)

(x − 1

limx→+∞

(x − 1

=(1+∞

)= lim

x→+∞

(x−1x

= limx→+∞

(1− 1

)x(1 + 1

e= e−2.

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

= limx→0

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

= limx→0

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

)= lim

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

)= lim

x· ln(1 + x)

= limx→0

(1 + x)1

= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

)= lim

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

)= lim

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(b)

ln(1 + x)

Rje²enje. Imamo

limx→0

ln(1 + x)

)= lim

x· ln(1 + x)

= limx→0

(1 + x)1

)= ln e

limx→0

ln(1 + x)

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

= limt→0

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 40(c)

ex − 1

Rje²enje. Imamo

limx→0

ex − 1

[t = ex − 1 ; ex = 1 + t ; x = ln(1 + t)

x → 0 ⇒ t → 0

]= lim

ln(1 + t)

= limt→0

1ln(1+t)

limx→0

ex − 1

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

= limx→0

ex−1x ·

sin xx ·

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

= limx→0

ex−1x ·

sin xx ·

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

)= lim

ex−1x · x

sin xx · x

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

)= lim

ex−1x ·�x

sin xx ·�x

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

)= lim

ex−1x ·�x

sin xx ·�x

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

)= lim

ex−1x ·�x

sin xx ·�x

= limx→0

ex−1x

sin xx

Zadatak 41(a)

ex − 1

sin x.

Rje²enje. Imamo

limx→0

ex − 1

sin x=

)= lim

ex−1x ·�x

sin xx ·�x

= limx→0

ex−1x

sin xx

Zadatak 41(b)

Za a ∈ 〈0,+∞〉, izra£unajte sljede¢i limes (ako postoji): limx→0

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

= limx→0

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

)x= ex ln a, x ∈ R.

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

= limx→0

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

= limx→0

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

= limt→0

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1

= ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(b)

ax − 1

Rje²enje. Imamo

limx→0

ax − 1

)= lim

ex ln a − 1

[t = x ln a ; x = t

ln ax → 0 ⇒ t → 0

]= lim

et − 1t

= limt→0

ln a · et − 1

t= ln a · 1 = ln a,

ax =(e ln a

Dakle,

limx→0

ax − 1

x= ln a.

Zadatak 41(c)

8x − 7x

6x − 5x.

Rje²enje. Imamo

limx→0

8x − 7x

6x − 5x=

= limx→0

8x−1x − 7x−1

x6x−1x − 5x−1

=ln 8− ln 7

ln 6− ln 5.

Zadatak 41(c)

8x − 7x

6x − 5x.

Rje²enje. Imamo

limx→0

8x − 7x

6x − 5x=

= limx→0

8x−1x − 7x−1

x6x−1x − 5x−1

=ln 8− ln 7

ln 6− ln 5.

Zadatak 41(c)

8x − 7x

6x − 5x.

Rje²enje. Imamo

limx→0

8x − 7x

6x − 5x=

)= lim

8x−1x − 7x−1

x6x−1x − 5x−1

=ln 8− ln 7

ln 6− ln 5.

Zadatak 41(c)

8x − 7x

6x − 5x.

Rje²enje. Imamo

limx→0

8x − 7x

6x − 5x=

)= lim

8x−1x − 7x−1

x6x−1x − 5x−1

=ln 8− ln 7

ln 6− ln 5.

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

= limx→0

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

= limx→0

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

= limt→0

(1 + t)1

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1

Zadatak 41(d)

(1 + sin x)1

Rje²enje. Imamo

limx→0

(1 + sin x)1

x =(1±∞

)= lim

((1 + sin x)

) sin xx.

Kako vrijedi

limx→0

(1 + sin x)1

sin x =

[t = sin x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

limx→0

slijedi da je

limx→0

(1 + sin x)1

x = e1 = e.

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

= limx→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

= limx→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) x+1

2(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) ��x+1

2��(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) ��x+1

2��(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) ��x+1

2��(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

= limt→0

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) ��x+1

2��(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(e)

(2x + 3

2x + 2

Rje²enje. Imamo

limx→+∞

(2x + 3

2x + 2

=(1+∞

)= lim

x→+∞

2x + 2

= limx→+∞

2x + 2

)2x+2) x+1

= limx→+∞

2x + 2

)2x+2) ��x+1

2��(x+1)

= limx→+∞

2x + 2

)2x+2) 1

[t = 1

2x+2; 2x + 2 = 1

x → +∞ ⇒ t → 0

]= lim

((1 + t)

Zadatak 41(f)

(1 + tg x)ctg x .

Rje²enje. Imamolimx→0

(1 + tg x)ctg x =(1±∞

= limx→0

(1 + tg x)1

[t = tg x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

Zadatak 41(f)

(1 + tg x)ctg x .

(1 + tg x)ctg x =(1±∞

= limx→0

(1 + tg x)1

[t = tg x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

Zadatak 41(f)

(1 + tg x)ctg x .

(1 + tg x)ctg x =(1±∞

)= lim

x→0(1 + tg x)

[t = tg x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

Zadatak 41(f)

(1 + tg x)ctg x .

(1 + tg x)ctg x =(1±∞

)= lim

x→0(1 + tg x)

[t = tg x

x → 0 ⇒ t → 0

= limt→0

(1 + t)1

Zadatak 41(f)

(1 + tg x)ctg x .

(1 + tg x)ctg x =(1±∞

)= lim

x→0(1 + tg x)

[t = tg x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

Zadatak 41(f)

(1 + tg x)ctg x .

(1 + tg x)ctg x =(1±∞

)= lim

x→0(1 + tg x)

[t = tg x

x → 0 ⇒ t → 0

]= lim

t→0(1 + t)

Zadatak 41(g)

log(1 + 10x)

Rje²enje. Koriste¢i da vrijedi

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

Rje²enje.

Koriste¢i da vrijedi

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

= limx→0

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

= limx→0

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

= limx→0

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

= limt→0

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

= 1 · 10

ln 10=

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10.

Zadatak 41(g)

log(1 + 10x)

loga b =logc b

logc a, a, c ∈ 〈0,+∞〉 \ {1} , b ∈ 〈0,+∞〉 ,

ra£unamo

limx→0

log(1 + 10x)

)= lim

ln(1+10x)ln 10

x= lim

ln(1 + 10x)

[t = 10x ; x = t

x → 0 ⇒ t → 0

]= lim

ln(1 + t)t10

= limt→0

ln(1 + t)

t· 10

ln 10= 1 · 10

ln 10=

ln 10.

Zadatak 42(a) (racionalizacija)

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

= limx→0

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

1 + sin x +√1− sin x

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

= limx→0

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

)= lim

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

)= lim

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

= limx→0

2 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

)= lim

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

)= lim

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

√1 + sin x −

√1− sin x

Rje²enje. Imamo

limx→0

√1 + sin x −

√1− sin x

)= lim

√1 + sin x −

√1− sin x

x·√1 + sin x +

√1− sin x√

= limx→0

(1 + sin x)− (1− sin x)

)= lim

x→02 · sin x

x· 1√

= 2 · 1 · 1√1 + sin 0 +

√1− sin 0

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· (√x + 1 + 1

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· (√x + 1 + 1

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· (√x + 1 + 1

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· (√x + 1 + 1

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

x + 1 + 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· �x

�x(√

x + 1 + 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· �x

�x(√

x + 1 + 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· �x

�x(√

x + 1 + 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(b)

2 sin(√

x + 1− 1)

Rje²enje. Imamo

limx→0

2 sin(√

x + 1− 1)

)= lim

x→02 ·

sin(√

x + 1− 1)

√x + 1− 1

·√x + 1− 1

x·√x + 1 + 1√x + 1 + 1

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· �x

�x(√

x + 1 + 1)

= limx→0

2 ·sin(√

x + 1− 1)

√x + 1− 1

· 1√x + 1 + 1

= 2 · 1 · 1√0 + 1 + 1

Zadatak 42(c)

Za a ∈ 〈0,+∞〉 i n ∈ N, izra£unajte sljede¢i limes (ako postoji): limx→a

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

= limx→a

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

((n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

= limx→a

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

((n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

((n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

((n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

n√x − n√a(

n√x − n√a) ((

n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

��n√x − n√a

��(n√x − n√a) ((

n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje.

Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

��n√x − n√a

��(n√x − n√a) ((

n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(c)

n√x − n√a

x − a.

Rje²enje. Koriste¢i da je

an − bn = (a− b)(an−1 + an−2b + . . .+ bn−1

), a, b ∈ R, n ∈ N,

ra£unamo

limx→a

n√x − n√a

x − a=

)= lim

n√x − n√a(

n√x)n − ( n

√a)n

= limx→a

��n√x − n√a

��(n√x − n√a) ((

n√x)n−1

n√x)n−2 · n

√a + . . .+

(n√a)n−1)

n√a)n−1

n√a)n−2 · n

√a + . . .+

(n√a)n−1

n√a)n−1 .

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

= limx→1

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(√x2 + 3x + 2x

) = limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

= limx→1

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(√x2 + 3x + 2x

) = limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(√x2 + 3x + 2x

) = limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(√x2 + 3x + 2x

) = limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(√x2 + 3x + 2x

) = limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x(x − 1)

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x��(x − 1)

��(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x��(x − 1)

��(x − 1)(√

x2 + 3x + 2x) = lim

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x��(x − 1)

��(x − 1)(√

x2 + 3x + 2x) = lim

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1

= −3

Zadatak 42(d)

√x2 + 3x − 2x

x − 1.

Rje²enje. Imamo

limx→1

√x2 + 3x − 2x

x − 1=

)= lim

√x2 + 3x − 2x

x − 1·√x2 + 3x + 2x√x2 + 3x + 2x

= limx→1

x2 + 3x − 4x2

(x − 1)(√

x2 + 3x + 2x) = lim

−3x2 + 3x

(x − 1)(√

x2 + 3x + 2x)

= limx→1

−3x��(x − 1)

��(x − 1)(√

x2 + 3x + 2x) = lim

−3x√x2 + 3x + 2x

=−3 · 1√

12 + 3 · 1 + 2 · 1= −3

Zadatak 42(e)

Izra£unajte sljede¢i limes (ako postoji): limx→−∞

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

Kako za sve x ∈ 〈−∞, 0〉 vrijedi x = −√x2, imamo

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

= −√

1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

Rje²enje. Imamolim

x→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

= limx→−∞

(√x2 − 5x + 4 + x

)·√x2 − 5x + 4− x√x2 − 5x + 4− x

= limx→−∞

−5x + 4√x2 − 5x + 4− x

·1x1x

= limx→−∞

−5 + 4x√

x2−5x+4x − 1

√x2 − 5x + 4

−√x2

= −√

x2 − 5x + 4

x2= −

√1− 5

x2, x ∈ 〈−∞, 0〉 ,

limx→−∞

(√x2 − 5x + 4 + x

)= lim

x→−∞

−5 + 4x

−√

1− 5x + 4

x2− 1

=−5 + 0

−√1− 0 + 0− 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

pri £emu sedma jednakost vrijedi jer je t =√t2 za sve t ∈ [0,+∞〉.

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

= limt→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

= limt→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

= limt→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1

Zadatak 42(e)

(√x2 − 5x + 4 + x

limx→−∞

(√x2 − 5x + 4 + x

)= ((+∞) + (−∞))

[t = −x ; x = −t

x → −∞ ⇒ t → +∞

]= lim

t→+∞

(√t2 + 5t + 4− t

)= lim

t→+∞

(√t2 + 5t + 4− t

)·√t2 + 5t + 4 + t√t2 + 5t + 4 + t

= limt→+∞

5t + 4√t2 + 5t + 4 + t

·1t1t

= limt→+∞

5 + 4t√

t2+5t+4t + 1

= limt→+∞

5 + 4t√

t2+5t+4√t2

= limt→+∞

5 + 4t√

t2+5t+4t2

+ 1= lim

t→+∞

5 + 4t√

1 + 5t + 4

=5 + 0√

1 + 0 + 0 + 1=

Oprez!

limx→0

x6= lim

x= lim

x→00 = 0.

De nicijaszunar/m1_materijali...sinx ne postoji. Primjer 1(f) sin 1 x 1 1 x y lim x!0 + sin 1 x i...

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