MATH20101, Real Analysis, Exam 2006-2007

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1) Answer all parts.

(a) By verifying the appropriate definitions prove

(i) limx→2

(2x2 − x + 1) = 7,

(ii) limx→∞

1

x2 + x + 1= 0.

(b) Suppose that f, g and h are three functions such that

h (x) ≤ f (x) ≤ g (x)

for all x in some deleted neighbourhood of a ∈ R.

Prove that if limx→a h (x) = limx→a g (x) = L then limx→a f (x) = L.

(c) Calculate

limx→0

x2 sin(

1x

)2− cos x

.

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Solution to Question 1 (a) (i) Let ε > 0 be given. Choose δ = min (1, ε/9) >0. Assume 0 < |x− 2| < δ. Then |x− 2| < δ ≤ 1 so 1 < x < 3 and thus|2x + 3| < 9. Therefore∣∣(2x2 − x + 1

)− 7

∣∣ =∣∣2x2 − x− 6

∣∣ = |(2x + 3) (x− 2)|

< 9 |x− 2| ≤ 9 (ε/9) = ε.

Hence we have verified the definition that limx→2 (2x2 − x + 1) = 7.Similar to problem on problem sheet [5 marks]

(ii) Let ε > 0 be given. Choose X = 1/ε > 0. Assume x > X. Then

1

x2 + x + 1<

1

xon “throwing away” x2 + 1 > 0,

<1

X= ε.

Hence we have verified the definition that limx→∞

1/ (x2 + x + 1) = 0.

Similar to problem on problem sheet [5 marks]

(b) There exists δ0 > 0 such that if 0 < |x− a| < δ0 then h (x) ≤ f (x) ≤g (x) .

Let ε > 0 be given.

The definition of limx→a h (x) = L implies there exists δ1 > 0 such thatif 0 < |x− a| < δ1 then |h (x)− L| < ε/3.

The definition of limx→a g (x) = L implies there exists δ2 > 0 such that if0 < |x− a| < δ2 then |g (x)− L| < ε/3.

Let δ = min (δ0, δ1, δ2) .

Assume x satisfies 0 < |x− a| < δ. For such x all h (x) ≤ f (x) ≤ g (x),|h (x)− L| < ε/3 and |g (x)− L| < ε/3 hold.

Consider

|f (x)− L| = |f (x)− h (x) + h (x)− L| ≤ |f (x)− h (x)|+ |h (x)− L|by the triangle inequality. The assumption h (x) ≤ f (x) ≤ g (x) implies0 ≤ f (x)−h (x) ≤ g (x)−h (x) . Since the differences are both positive theyequal their moduli, and so |f (x)− h (x)| ≤ |g (x)− h (x)|. Thus

|f (x)− h (x)|+ |h (x)− L| ≤ |g (x)− h (x)|+ |h (x)− L|

= |g (x)− L + L− h (x)|+ |h (x)− L|

≤ |g (x)− L|+ |L− h (x)|+ |h (x)− L|

by the triangle inequality again,

< ε/3 + ε/3 + ε/3 = ε.

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Hence we have verified the definition that limx→a f (x) = L.Bookwork [6 marks]

(c) Using −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1 we have for x 6= 0,

−x2 ≤ x2 sin

(1

x

)≤ x2,

since x2 > 0 and 1 ≤ 2− cos x ≤ 3. Hence

−x2 = − x2

2− cos x<

x2 sin(

1x

)2− cos x

<x2

2− cos x= x2.

Since limx→0 x2 = 0 the sandwich rule gives

limx→0

x2 sin(

1x

)2− cos x

= 0.

[4 marks]

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2) Answer all parts.

(a) (i) If limx→a g (x) = L exists and f is continuous at L prove that

limx→a

f (g (x)) = f(

limx→a

g (x))

= f (L) .

(ii) Show that

cos

(x− 2

x2 − 4x + 8

)is continuous on R.

All standard results on continuity you assume should be carefully stated.

(b) Carefully state the Intermediate Value Theorem.

Prove thatx

sin x+

1

cos x= π

has a solution with x ∈ (0, π/2).

Prove that the solution found is unique within (0, π/2) .

State carefully any results you use to prove this. It may be of use toassume that tan x > x in (0, π/2).

4

Solution to Question 2 (a) (i) Let ε > 0 be given. f is continuous at Limplies there exists δ1 such that

if |y − L| < δ1 then |f (y)− f (L)| < ε. (1)

Apply the definition of limx→a g (x) = L with ε replaced by δ1 to findδ > 0 such that

if 0 < |x− a| < δ then |g (x)− L| < δ1. (2)

Combine (2) and (1) with y = g (x) to get

if 0 < |x− a| < δ then |f (g (x))− f (L)| < ε.

Thus we have verified the definition that limx→a f (g (x)) = f (L) .Bookwork [6 marks]

(ii) We assume the results that rational functions are continuous at allpoints at which they are defined while cos is continuous on R.

We need check that (x− 2) / (x2 − 4x + 8) is defined on R.

Either note that x2 − 4x + 8 = (x− 2)2 + 4 > 0 and so never zero, ornote that because (−4)2 − 4 × 8 < 0 the roots are complex and so, again,x2 − 4x + 8 is never zero for x ∈ R.

Apply part (i). [3 marks]

(b) Intermediate Value Theorem. Suppose that f is a function continuouson a closed interval [a, b] and that f (a) 6= f (b). If γ is some number betweenf (a) and f (b) then there must be at least one c : a < c < b for whichf (c) = γ.

Bookwork [2 marks]

Let f (x) = x/ sin x + 1/ cos x. To use the Intermediate Value Theoremwe need a closed interval, which cannot be [0, π/2] since f is not defined atthe end points. Instead we look at points within this interval. We find that

f(π

6

)=

2π

6+

2√3≈ 2.20 < π,

f(π

3

)=

2π

3√

3+ 2 ≈ 3.209 > π.

Hence, by applying the Intermediate Value Theorem with γ = π on[π/6, π/3] , we find there is a solution to f (x) = π somewhere in (0, π/2).

Similar to problem covered in lectures [5 marks]

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Assume for contradiction that there is more than one solution in (0, π/2),i.e. there exist a, b ∈ (0, π/2) such that f (a) = π = f (b) . We then havean example of a function continuous on [a, b], differentiable on (a, b) withf (a) = f (b). These are exactly the conditions of Rolle’s Theorem whichimplies the existence of c ∈ (a, b) ⊆ (0, π/2) with f ′ (c) = 0. Yet

f ′ (c) =sin c− c cos c

sin2 c+

sin c

cos2

=cos c (tan c− c)

sin2 +sin c

cos2

> 0

by the assumption given in the question. This contradiction proves theuniqueness of solution.

Similar to problem covered in lectures [4 marks]

6

3) Answer all parts.

(a) State the Mean Value Theorem.

Prove

(i) sin x < x for x > 0 and (ii) ln (1 + x) < x for x > 0.

Deducee−x sin x <

x

1 + x

for x > 0.

(b) Let f, g : A → R both be differentiable at a ∈ A. Prove, using therules of limits, that

(fg)′ (a) = f ′ (a) g (a) + f (a) g′ (a) .

(c) Calculate the Taylor polynomial

T7,0

(e−x sin x

).

Prove, using Lagrange’s form for the error, that∣∣e−x sin x− T7,0

(e−x sin x

)∣∣ ≤ 3.61× 10−13

for |x| ≤ 10−1.

Hint, You should use the result from Part a in bounding the result thatarises from the application of Lagrange’s form of the error.

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Solution to Question 3 (a) Mean Value Theorem. Let f : [a, b] → R becontinuous on [a, b] and differentiable on (a, b) . Then there exists c ∈ (a, b)such that

f (b)− f (a)

b− a= f ′ (c) .

Bookwork [2 marks](i) With f (x) = sin x we get, for some 0 < c < x,

sin x− sin 0

x− 0= cos c ≤ 1.

So for x > 0 we can multiply up to get result. From Lectures [1 mark]

(ii) With f (x) = ln (1 + x) we get, for some 0 < c < x,

ln (1 + x)− 0

x− 0=

1

1 + c< 1.

So for x > 0 we can multiply up to get result. From Lectures [1 mark]

Exponentiating ln (1 + x) < x we get 1 + x < ex or e−x < 1/ (1 + x).Combine with (i) to get

e−x sin x <x

1 + x

for x > 0. [1 mark]

(b) Consider

f (x) g (x)− f (a) g (a)

x− a=

f (x) g (x)− f (a) g (x) + f (a) g (x)− f (a) g (a)

x− a

= g (x)f (x)− f (a)

x− a+ f (a)

g (x)− g (a)

x− a.

Then by the sum and product rules for limits, allowable since the finallimits exist, and the fact that a function is continuous when differentiable,

(fg)′ (a) = limx→a

g (x) limx→a

f (x)− f (a)

x− a+ f (a) lim

x→a

g (x)− g (a)

x− a

= g (a) f ′ (a) + f (a) g′ (a)

as required. Bookwork [4 marks]

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(c) Let f (x) = e−x sin x. Then f (0) = 0 while

f ′ (x) = −e−x sin x + e−x cos x, so f ′ (0) = 1,

f ′′ (x) = e−x sin x− e−x cos x− e−x cos x− e−x sin x

= −2e−x cos x, so f ′′ (0) = −2,

f (3) (x) = 2e−x cos x + 2e−x sin x, so f (3) (0) = 2,

f (4) (x) = −2e−x cos x− 2e−x sin x− 2e−x sin x + 2e−x cos x

= −4e−x sin x = −4f (x) , so f (4) (0) = 0.

The process now repeats so

f (5) (x) = −4f ′ (x) , so f (5) (0) = −4,

f (6) (x) = −4f ′′ (x) , so f (6) (0) = 8,

f (7) (x) = −4f (3) (x) , so f (7) (0) = −8,

and f (8) (x) = −4f (4) (x) = 16f (x). Hence

T7,0

(e−x sin x

)= 0 + x + (−2)

x2

2!+ 2

x3

3!+ 0

x4

4!+ (−4)

x5

5!+ 8

x6

6!+ (−8)

x7

7!

= x− x2 +x3

3− x5

30+

x6

90− x7

630.

Similar to problem covered in lectures [8 marks]

Using Lagrange’s form of the error we have, for some c between 0 and x,

∣∣e−x sin x− T7,0

(e−x sin x

)∣∣ =

∣∣f (8) (c)∣∣

8!x8 =

16 |f (c)|8!

x8

=16e−c |sin c|

8!x8

≤ 16

8!

c

1 + cx8 by part (a),

≤ 16

8!

1

1110−8 since c < x < 10−1

< 3.61× 10−13

using a calculator. Similar to problem covered in lectures [3 marks]

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4) Answer all parts.

(a) Calculate the lower and upper sums

L (Qn, f) and U (Qn, f) ,

where f : [2, 4] , x → 2x2, and

Qn ={2ηi : 0 ≤ i ≤ n

}is the n-th geometric partition of [2, 4], so η = 21/n.

(b) Let f : [a, b] → R be a bounded function.

Define the lower and upper integrals,∫ b

a

f and

∫ b

a

f .

(c) Assuming that L (P , f) ≤ U (D, f) for all partitions P and D of [a, b],prove ∫ b

a

f ≤∫ b

a

f .

Define what is meant by saying that f : [a, b] → R, a bounded function,is Riemann Integrable.

(d) Prove that f : [2, 4] , x → 2x2 is Riemann integrable and find thevalue of the integral.

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Solution to Question 4 (a) f (x) = 2x2 is an increasing function on [2, 4]

so on each subinterval [xi−1, xi] = [2ηi−1, 2ηi] we have mi = 2(2η(i−1)

)2and

Mi = 2 (2ηi)2. Then

U (Qn, f) =n∑

i=1

Mi (xi − xi−1)

=n∑

i=1

2(2ηi

)2 (2ηi − 2ηi−1

)= 16

(1− η−1

) n∑i=1

η3i using ηi − ηi−1 = ηi(1− η−1

)= 16

(1− η−1

) η3 (η3n − 1)

η3 − 1on summing the geometric series,

= 16 (η − 1)η2 (23 − 1)

(η − 1) (η2 + η + 1)since ηn = 2

=112η2

η2 + η + 1.

We don’t wish to do the same work twice so we relate the Lower sum to theUpper one by

L (Qn, f) =n∑

i=1

mi (xi − xi−1)

=n∑

i=1

2(2ηi−1

)2 (2ηi − 2ηi−1

)=

1

η2U (Qn, f) =

112

η2 + η + 1.

Similar to problem covered in lectures [8 marks]

(b) For f bounded on [a, b], the Upper Integral is∫ b

a

f = glb {U (P , f) : P a partition of [a, b]}

and the Lower Integral is∫ b

a

f = lub {L (P , f) : P a partition of [a, b]} .

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Bookwork [2 marks]

(c) By assumption,L (P , f) ≤ U (D, f)

for all partitions P and D. Fix D and vary P . We see that U (D, f) is

an upper bound for {L (P , f) : P}. But∫ b

af is, by definition, the least of all

upper bounds, hence U (D, f) ≥∫ b

af .

Now vary D and see that∫ b

af is a lower bound for {U (D, f) : D}. But∫ b

af is, by definition, the greatest of all lower bounds, so

∫ b

a

f ≤∫ b

a

f

as required. Bookwork [5 marks]

(d) For all partitions we have

L (P , f) ≤∫ 4

2

f ≤∫ 4

2

f ≤ U (P , f) .

Taking the sequence of Qn from part (a) we see that

112

η2 + η + 1≤

∫ 4

2

f ≤∫ 4

2

f ≤ 112η2

η2 + η + 1.

Let n →∞ when η → 1 to get

112

3≤

∫ 4

2

f ≤∫ 4

2

f ≤ 112

3.

Thus we must have equality throughout, so the lower integral equals theupper and hence f in integrable over [2, 4]. The common value of 112/3 isthen the value of the integral.

Similar to problem covered in lectures [5 marks]

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