CHAPTER 14. PARTIAL DERIVATIVES 83

14.2 Limits and Continuity

Comments. Recall that the Calculus 1 definition of limit, says, roughly,

limx!a

f(x) = Lmeans that the y-values of f(x) become infinitelyclose to L as we make the x-values infinitely closeto the number a.

We can make this definition more precise as by replacing “close” with the distancebetween two numbers:

limx!a

f(x) = Lmeans that we can make |f(x)�L| as close to 0 aswe want, by plugging in x values that make |x�a|su�ciently close to 0.

Definition. Let f be a function of (x, y) and let a = ha, bi be in the domain of f . Letx = hx, yi. Then

limx!a

f(x, y) = L meanswe can make |f(x)� L| as close to 0 as we want byplugging in x values that make |x� a| su�cientlyclose to 0.

There are other notations for this limit:

f(x, y) !L as (x, y) ! (a, b)

limhx, yi!ha, bi

f(x) = L (where x = hx, yi and a = ha, bi)

Comments. There is some ambiguity in the above definition (this is just the informalversion: the formal version has no ambiguity). The main issue is this: which values ofhx, yi do we need to consider plugging in to make f(x, y) close to L? Some values thatare close to ha, bi? All values that are close to ha, bi?

The most conservative answer is to require that f(x, y) be close to L for all values ofhx, yi that are close to ha, bi. In particular, we have to require the following: no matterhow we pick values for (x, y) that are approaching (a, b) we must get the same limit forf(x, y) to say that lim f(x, y) = L.

In particular suppose we have two paths, C1 and C2 that are both approaching thepoint (a, b). If f(x, y) ! L1 as (x, y) ! (a, b) along path C1 and f(x, y) ! L2 as(x, y) ! (a, b) along path C2, then the limit DNE.

Example 1. Explore the following limits graphically and algebraically.

(a) lim(x,y)!(0,0)

x2 + y2.

(b) lim(x,y)!h�2, 3i

(x3 � 2yx2 + y2 � 5xy3)

Solution:

(a) The two pictures below show the graphs of z = x2 + y2, one with a larger domain,and one with a smaller one.

CHAPTER 14. PARTIAL DERIVATIVES 84

04

10

2 4

20z

z=x2+y2

2

30

y

0

x

40

0-2

-2-4 -4

-0.10.1

-0.05

0.05 0.1

0z

z=x2+y2

0.05

0.05

y

0

x

0.1

0-0.05

-0.05-0.1 -0.1

[x,y]= meshgrid(linspace (-4,4,35));

z=x.^2+y.^2;

mesh(x,y,z)

xlabel('x'),ylabel('y'),zlabel('z')title('z=x^2+y^2')

We can see that the z-values are pretty small, in fact they are close to 0. Alge-braically, we would guess that maybe we can just plug the numbers in. If we dothis we get

limhx, yi!h0, 0i

x2 + y2 = 02 + 02 = 0

and from the graph this looks correct.

(b) This time let’s try a small leap of faith, namely that the algebra works, and thenverify our answer with a graph. If we do this we get:

limhx, yi!h�2, 3i

x3 � 2yx2 + y2 � 5xy3 = (�2)3 � 2(3)(�2)2 + (3)2 � 5(�2)(3)3

= �8� 24 + 9 + 270

= 247

Now we double check with a graph. The pictures below show the graphs of z =x3 � 2yx2 + y2 � 5xy3, one with a larger domain, and one with a smaller one.

-610

-4

-2

5 10

0z

104

2

Example: x 3-2yx2+y2-5xy3

limit as (x,y) -> (-2,3)

5

y

0

4

x

6

0-5 -5

-10 -10−2.3

−2.2−2.1

−2−1.9

−1.8−1.7

−1.6

2.6

2.8

3

3.2

3.4150

200

250

300

350

400

x

Example: limit as (x,y) −> (−2,3)

y

z

CHAPTER 14. PARTIAL DERIVATIVES 85

u=linspace (-4,0,25);

v=linspace (1,5,25);

[x,y]= meshgrid(u,v);

z=x.^3-2*y.*x.^2+y.^2 -5.*x.*y.^3;

mesh(x,y,z)

xlabel('x'),ylabel('y'),zlabel('z')title('Example: limit as (x,y) -> (-2,3)')

Even the first graph suggests that we were correct in our algebraic leap of faith.The z-values are pretty large, certainly in the hundreds. The second graph showsthe surface zoomed in to �2.3 x �1.6 and 2.6 y 3.4. If you trace thevalues of x = �2 and y = 3 on the together on the graph, you can tell that z isjust a little below 250.

Comments. The Limit Laws from section 2.3 all apply to the limits we are consideringhere. In addition, the Squeeze Theorem applies. One result from Calculus I that doesnot apply is L’Hospital’s Rule, at least in general. In other words, there is no way togeneralize L’Hospital’s Rule to most multivariable limits.

Definition. Let f be a function of several variables.

f is continuous at ha, bi means limhx, yi!ha, bi

= f(a, b).

Theorem. Let f be a function of several variables. If f is any finite combination ofpowers of x, ex, ln(x), trig functions or inverse trig functions, then f is continuous ateach point in its domain. In other words, we can evaluate a limit of f by plugging in,provided the result is defined.

Comments. Note: Although many basic facts about limits in Calculus I can be ex-tended to the multivariable case, there is one important result that does not have anysimple extension: L’Hospital’s Rule.

Example 2. Explore the following limits graphically. What happens if you try to thesealgebraically? Can you figure out for sure what’s going on?

(a) limhx, yi!h0, 0i

xy

x2 + y2

Solution: We show below two views of the first graph

−4

−2

0

2

4

−4

−2

0

2

4−0.5

0

0.5

x

g(x,y)=(xy)/(x2+y2)

y

z

−4−3−2−101234

−4

−2

0

2

4

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

y

x

g(x,y)=(xy)/(x2+y2)

z

CHAPTER 14. PARTIAL DERIVATIVES 86

[x,y]= meshgrid(linspace (-4,4,35));

z=x.*y./(x.^2+y.^2);

mesh(x,y,z)

xlabel('x'),ylabel('y'),zlabel('z')title('g(x,y)=(xy)/(x^2+y^2)')

Maybe you can see that it looks like two di↵erent things are happening: some ofthe z-values are approaching 1/2 and some are approaching �1/2. In fact, it’seven possible to figure out two di↵erent paths that produce these results.

Consider the limit taken along the path y = x: f(x, y) =x2

x2 + x2=

x2

2x2=

1

2.

Thus, along this path, the limit equals 1/2.

Now consider a di↵erent path: y = �x: f(x, y) =�x2

x2 + x2=

�x2

2x2= �1

2. Thus,

along this path, the limit equals �1/2.

Since one path to h0, 0i gives a limit of 1/2 and one path gives a limit of �1/2,the limit does not exist, which confirms what we saw in the graph.

(b) limhx, yi!h0, 0i

sin⇣p

x2 + y2⌘

px2 + y2

This is where we ended on Wednesday, February 13

Solution: We show below two views of the second graph: the di↵erence is the sizeof the domain

[x,y] = meshgrid(linspace (-10,10,40));

R=sqrt(x.^2 + y.^2);

z = sin(R)./R;

mesh(x,y,z)

title ('Example 3 -- the sinc function ')xlabel('x'), ylabel('y'), zlabel('f(x,y)')

−10

−5

0

5

10

−10

−5

0

5

10−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

Example 3 −− the sinc function

y

f(x,

y)

−1

−0.5

0

0.5

1

−1

−0.5

0

0.5

10.65

0.7

0.75

0.8

0.85

0.9

0.95

1

x

Example 3 −− the sinc function

y

f(x,

y)

It appears that the limit equals 1. In fact, this can be proven as follows.

CHAPTER 14. PARTIAL DERIVATIVES 87

Let u =px2 + y2. Then

limx!0

u = limx!0

px2 + y2 =

p0 = 0.

Thus

limx!0

sin(px2 + y2)p

x2 + y2== lim

x!0

sin(u)

u= lim

u!0

sin(u)

u= 1

Example 3. Find the following limit algebraically

lim(x,y)!(0,0)

x3 + xy2

x2 + y2

Solution:

limx(x2 + y2)

x2 + y2= limx = 0

Example 4. Apply the Squeeze Theorem to find the following limit:

limhx, yi!h0, 0i

x2 sin2(y)

x2 + 2y2

Solution: We start by getting a pair of inequalities that we can work with:

0 y2 and 0 x2 squaring numbers makes them � 0

0 2y2 multiply previous by 2

x2 x2 + 2y2 add x2 to previous

0 x2 x2 + 2y2 combine previous with 1st line

0 x2

x2 + 2y2 1 divide previous by x2 + 2y2

0 x2 sin2(y)

x2 + 2y2 sin2(y) multiply previous by sin2(y)

limhx, yi!h0, 0i

0 = 0

limhx, yi!h0, 0i

sin(y2) = sin(0)

= 0

Therefore, by the Squeeze Theorem, we have

limhx, yi!h0, 0i

x2 sin2(y)

x2 + 2y2= 0

Exploring limits

Example 5. In the definition of limit given above, we vaguely described that the limithas to converge for all hx, yi near ha, bi. However, we often explored those situationswhere this convergence didn’t happen for hx, yi following di↵erent paths to ha, bi, typ-ically straight lines. This raises various questions.

CHAPTER 14. PARTIAL DERIVATIVES 88

(a) T/F: If limhx, yi!ha, bi

f(x, y) = L, then limhx, yi ! ha, bi

y = x

f(x, y) = L.

(b) T/F: If limhx, yi!ha, bi

f(x, y) = L, then limhx, yi ! ha, bi

y = mx, for all m

f(x, y) = L.

(c) T/F: If limhx, yi ! ha, bi

y = mx, for all m

f(x, y) = L, then limhx, yi!ha, bi

f(x, y) = L.

(d) T/F: If limhx, yi!ha, bi

f(x, y) = L, then limhx, yi ! ha, bi

x = f(t), y = g(t), for all paths

f(x, y) =

L.

(e) T/F: If limhx, yi ! ha, bi

x = f(t), y = g(t), for all paths

f(x, y) = L, then limhx, yi!ha, bi

f(x, y) =

L.

Solution:

(a) T

(b) T

(c) F. See Fleming, Functions of Several Variables, 2.2, example 2 or https://www.maa.org/sites/default/files/nanyes09200535167.pdf or Apostol’s Calculus,Vol II, Exercise 8, p. 165.

(d) T.

(e) T.

Further questions arise: as we see from part (c), the limit existing along every linedoes not prove that the limit exists in full, but we see from part (e) that if the limitexists along all curves, then the limit exists in full.

One might ask: what if the limits exist (and are all equal) along all powers of x, i.e.paths of the form mxn for various m and n. This is insu�cient too. What about allpaths formed by polynomials? My guess, and it’s just a guess, is that this is su�cient.Why? Because polynomials are dense, meaning every path can by approximated by apolynomial to an arbitrary degree of accuracy. But again, although the last sentence istrue, it’s just a guess that this implies the sentence before that.