CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES

1. Definitions

1.1. Limit. Let f : Rn → Rm some function, x0 = (x1, . . . , xn) ∈ Rn and y0 =(y1, . . . , ym) ∈ Rm. Then

limx→x0

f(x) = y0

if and only for x ”close to” x0, f(x) is ”close to” y0. In other words, given ε > 0,there exists δ depending on ε and x0 (so we write δ = δ(ε,x0)) such that

dRn(x,x0) < δ ⇒ dRm(f(x),y0) < ε

where the distance on the left is taken in Rn, whereas the one on the right-handside is taken in the target space Rm. We will drop the subscript Rn from d, andwe will often write ‖x − x0‖ instead of d(x,x0). We will also drop the bold-facenotation.

Note: a function f : Rn → Rm is clearly given by a row vector f = (f1, . . . , fm)where fi’s are the components of f . In fact, fi = pi ◦ f (see below). Thenlimx→x0 f(x) = y0 if and only if limx→x0 fi(x) = yi, i = 1, . . . ,m.

1.2. Continuity. A function f : Rn → Rm is continuous at x0 ∈ Rn iff

limx→x0

f(x) = f(x0)

We say that f is continuous (everywhere) if it is continuous at every point of thedomain Rn.

Note: if f = (f1, . . . , fm), where f1, . . . , fm are the components of f , then f iscontinuous iff f1, . . . , fm are continuous, as functions : Rn → R.

2. Tools

Operations with limits: addition, subtraction, multiplication, division (when pos-sible), etc.

Components. If f : Rn → Rm, f = (f1, . . . , fm), then limx→x0 f(x) = y0 =(y1, . . . , ym) iff limx→x0 fi(x) = yi, for every i = 1, . . . ,m.

Projections. The functions pi : Rn → R, pi(x1, . . . , xn) = xi are continuous, andthis can be easily checked with the (ε, δ)- definition of continuity.

Composition of functions. If f : Rn → Rm is continuous and g : Rm → RN iscontinuous, then g ◦ f : Rn → RN is continuous, where g ◦ f(x) := g(f(x)).

Rn f−−−−→ Rm g−−−−→ RN

1

2 CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES

3. Examples

1. f : R3 → R, f(x, y, z) = x sin(z).Then f = p1 · (sin ◦p3) = continuous.

2. f : R3 → R2, f(x, y, z) = (xy, x + y + z2).• f1(x, y) = xy = p1 · p2 = cont.• f2(x, y) = x + y + z2 = p1 + p2 + (p3)2 = cont.

Therefore f = (f1, f2) is continuous.

3. f : R2 → R, f(x, y) =

{x2√

x2+y2, (x, y) 6= (0, 0)

0, (x, y) = (0, 0)Proof that f is continuous.

On R2 − (0, 0), f is continuous since it is a ratio of continuous functions (withnon-vanishing denominator), namely f = p2

1√·◦(p2

1+p22)

= continuous.NB: To make sure we understand why is the denominator continuous, we can writethe denominator as the composition of functions

R2 p21+p2

2−−−−→ [0,∞)√·−−−−→ R

We are thus left to show that f is continuous at (0, 0), in other words to prove that

lim(x,y)→(0,0)

x2√x2 + y2

= 0

Squeeze:

0 ≤ x2√x2 + y2

≤ x2

√x2

= |x|

However, as (x, y) → (0, 0) necessarily x → 0 and hence |x| → 0, therefore by thePinching Lemma x2√

x2+y2→ 0 as (x, y) → 0.

Different way of saying it ((ε, δ)-proof). Note that

d((x, y), (0, 0)) ≤ ε ⇔√

x2 + y2 ≤ ε ⇒ |x| ≤ ε ⇒ x2√x2 + y2

≤ |x| ≤ ε

4. f : R2 → R, f(x, y) =

{xy

x2+y2 , (x, y) 6= (0, 0)0, (x, y) = (0, 0)

.

Proof that f is not continuous at (0, 0). Consider the followin sequences of pointsin R2:

xk = (1k

, 0), k ≥ 1

yk = (1k

,1k

), k ≥ 1

Then xk → (0, 0) and yk → (0, 0) as k →∞, and yet

f(xk) = 0 → 0, k →∞

f(yk) =12→ 1

2, k →∞

which shows that the limitlim

x→(0,0)f(x)

does not exist. In particular, the function f is not continuous at (0, 0).

CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES 3

5. Let

g(x, y) =

{x sin(1/y), y 6= 00, y = 0

Determine the points (x, y) ∈ R2 where g is continuous.

6. Tricky example. Let f : R2 → R defined by

f(x, y) =

{xy2

x2+y4 , (x, y) 6= (0, 0)0, (x, y) = (0, 0)

Show that limk→∞ f(xk) = 0, for any sequence xk → (0, 0) such that xk, k ≥ 1 ison a fixed line passing through (0, 0). Is f continuous at (0, 0)?