14.2 1

14.2 Limits and Continuity

The following definitions should be familiar.

Definition. Limit of a Function of Two Variables

We say that f (x, y) approaches the limit L ∈ R as (x, y) approaches(x0, y0) and write

lim(x,y)→(x0,y0)

f (x, y) = L

if, for every ε > 0 there is a δ = δ(ε) > 0 such that for all (x, y) ∈ Dom(f )

0 <

√

(x− x0)2 + (y − y0)

2 < δ =⇒ |f (x, y)− L| < ε(1)

Remark. Notice that (1) can be rewritten using “vector norms” as

0 < ‖(x, y)− (x0, y0)‖ < δ =⇒ |f (x, y)− L| < ε

so that the definition above is identical to the one-dimensional versionfrom a first semester calculus course. That is, suppose that L ∈ R. Wesay that limx→c f (x) = L provided that for every ε > 0 there is a δ > 0

such that

0 < |x− c| < δ =⇒ |f (x)− L| < ε(2)

14.2 2

Using (2) we were able to deduce the following fundamental limits.

limx→c

k = k(3)

limx→c

x = c(4)

Combining the above results with the basic “Limit Laws” , we were ableto compute a large number of limits. For example, if we combine (3)and (3) with the Sum and Product Limit Laws, we can conclude that

limx→3

(

5x2 − 4x)

= 5(3)2 − 4(3)

We proceed along the same lines in this course. First, using (1) we caneasily prove that

lim(x,y)→(x0,y0)

x = x0(5)

lim(x,y)→(x0,y0)

y = y0(6)

lim(x,y)→(x0,y0)

k = k, k ∈ R(7)

Remark. For example, (6) implies that the function f (x, y) = y has alimit at every point (x0, y0) ∈ R

2.

Now we need the multivariable analogues of the Limit Laws.

14.2 3

Theorem 1. Properties of Limits - Multivariable Case

Let L, M, and k ∈ R and suppose that lim(x,y)→(x0,y0) f (x, y) = L andlim(x,y)→(x0,y0) g(x, y) = M .

1. Sum and Difference Law:

lim(x,y)→(x0,y0)

(f (x, y)± g(x, y)) = L±M

2. Product Law:

lim(x,y)→(x0,y0)

(f (x, y) · g(x, y)) = L ·M

3. Quotient Law:

lim(x,y)→(x0,y0)

f (x, y)

g(x, y)=

L

M, M 6= 0

4. Power Rule: If r, s ∈ Z with no common factors and s 6= 0 then

lim(x,y)→(x0,y0)

(f (x, y))r/s = Lr/s

If s is even then L > 0.

14.2 4

Example 1. Using the Limit Laws

Find the following limits, if they exist.

a.

lim(x,y)→(3,2)

(

5xy2 − 6y − 4)

= 5(3)(2)2 − 6(2)− 4

b.

lim(x,y)→(0,0)

xy − y2√x−√

y= lim

(x,y)→(0,0)

xy − y2√x−√

y

√x +

√y√

x +√y

= lim(x,y)→(0,0)

x (x− y)(√

x +√y)

x− y

= lim(x,y)→(0,0)

x(√

x +√y)

= 0

14.2 5

There is also a 2-dimensional analogue of the Squeeze Law.

Theorem 2. Suppose that for all (x, y) in some disk centered at(x0, y0) we have f (x, y) ≤ g(x, y) ≤ h(x, y), except possibly at (x0, y0). Inaddition, suppose that

lim(x,y)→(x0,y0)

f (x, y) = L = lim(x,y)→(x0,y0)

h(x, y)

Then

lim(x,y)→(x0,y0)

g(x, y) = L

Example 2. Prove that

(8) lim(x,y)→(0,0)

xy2

x2 + y2= 0

Suppose that (x, y) 6= (0, 0). Then 0 < y2 ≤ x2 + y2 and

y2

x2 + y2≤ 1

Thus∣

∣

∣

∣

xy2

x2 + y2

∣

∣

∣

∣

=|x|y2x2 + y2

≤ |x|

or

−|x| ≤ xy2

x2 + y2≤ |x|

And the result now follows by the Squeeze Law since

lim(x,y)→(0,0)

±|x| = 0

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Example 3. Functions without Limits

Let f (x, y) =3xy

x2 + y2. Evaluate the following limit or show that it does

not exist.lim

(x,y)→(0,0)f (x, y)

We first observe that the limit is of the (indeterminate) form, 0/0. Ourexperience in calculus I doesn’t seem to be much help here since theredoesn’t seem to be any obvious way to cancel the offending expressionfrom the denominator. However, if the limit does exist, then it must existalong any path to the origin (within the domain of f ).

Let’s try to compute the limit along any line to the origin. So let y = mx

for some m ∈ R. Then

lim(x,y)→(0,0)

3xy

x2 + y2= lim

x→0

3x(mx)

x2 + (mx)2

= limx→0

3m

1 +m2

x2

x2

= limx→0

3m

1 +m2

=3m

1 +m2

So the limit seems to depend on m. For example, the limit is 0 alongthe x-axis but the limit is 6/5 along the line y = 2x. In other words, theabove limit does not exist.

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The last example suggests the following two-dimensional analogue tothe “two-sided” limit theorem from calculus I.

Proposition 3. Two-Path Test for Nonexistence of a Limit

If a function f (x, y) has different limits as (x, y) approaches (x0, y0)

along two different paths then lim(x,y)→(x0,y0) f (x, y) does not exist.

Example 4. Functions without Limits - Part 2

Let f (x, y) =x2

y − x. Based on a sketch of the level curves, we suspect

that the following limit does not exist.

lim(x,y)→(0,0)

x2

y − x

Let’s try the same approach as we did in the previous example. So lety = mx for some m ∈ R \ {1} (Why is 1 excluded?). Then

lim(x,y)→(0,0)

x2

y − x= lim

x→0

x2

mx− x= lim

x→0

x

m− 1= 0

That’s not good. Now what? Let’s try the path y = mx2, for arbitrarym ∈ R.

lim(x,y)→(0,0)

x2

y − x= lim

x→0

x2

mx2 − x= lim

x→0

x

mx− 1= 0

This is frustrating.

14.2 8

Let’s take another look at the level curves of this function.

f(x, y) = −3

f(x, y) = 1/2

f(x, y) = 4

Now the approach seems clear. We try using an arbitrary level curveas the path to the origin. So let y = mx2 + x for some m ∈ R. Then

lim(x,y)→(0,0)

x2

y − x= lim

(x,y)→(0,0)

x2

mx2 + x− x

= limx→0

x2

mx2

= limx→0

1

m

= 1/m

Now since m was arbitrary we conclude that the limit does not exist.

14.2 9

Example 5. Evaluate the limit below or show that it does not exist.

(9) lim(x,y)→(0,0)

x3

x2 − y

The above limit is of the indeterminate form 0/0, so it could be

anything. Let f (x, y) =x3

x2 − y. Perhaps we might gain some insight if

we sketch the graph of z = f (x, y).

Figure 1: z =x3

x2 − y: Northeast Viewpoint

Figure 2: z =x3

x2 − y: Southeast Viewpoint

14.2 10

It is quite clear from (9) that f (x, y) is not defined on the parabolay = x2 and the above figures seem to confirm this. What now?

Let’s sketch a few level curves. So let f (x, y) = c and rearrange theresulting equation until we obtain something that we know how tograph.

x3

x2 − y= c

x3 = cx2 − cy

=⇒ y = x2(1− x/c)(10)

Now we sketch a few of these using values c = −3, 1, and 4.

f = 4f = −3

b P

bQ

Figure 3: Contour Map of f

Notice that the level curves happen to meet at the origin. So, forexample, if Q is a particle traveling along the level curve f (x, y) = 4,when the particle reaches the origin (or any other place on the curve) it

14.2 11

will always be at the “elevation” z = 4. In other words,

lim(x,y)→(0,0)

along f=4

f (x, y) = lim(x,y)→(0,0)

along f=4

4 = 4

Similarly

lim(x,y)→(0,0)

along f=−3

f (x, y) = lim(x,y)→(0,0)

along f=−3

−3 = −3

Since these two curves meet at the origin, the function does not have alimit there. The path is now clear.

Using (10) we compute the limit along an arbitrary level curve.

lim(x,x2−x3/c)→(0,0)

f (x, x2 − x3/c) = lim(x,x2−x3/c)→(0,0)

x3

x2 − (x2 − x3/c)

= limx→0

x3

x3/c

= limx→0

c = c

Now since c was arbitrary, the result follows by an application of thetwo-path test.

14.2 12

Continuity is defined in the usual way.

Definition. Continuity of a Multivariable Function

A function f (x, y) is continuous at (x0, y0) if

(11) lim(x,y)→(x0,y0)

f (x, y) = f (x0, y0)

A function is called continuous if it is continuous at each point of itsdomain.

Compare this with the continuity definition for real-valued functions of asingle variable.

It should be clear that constant functions, as well as, the functionsf (x, y) = x and g(x, y) = y are continuous at all points in the plane. Inaddition, the function h(x, y) = 5xy2 − 6y − 4 is continuous at (3, 2).

14.2 13

Finally, we have a very important composition theorem.

Theorem 4. The Composition of Continuous Functions isContinuous

If f is continuous at (x0, y0) and g is continuous at f (x0, y0) then thefunction h = g ◦ f defined by h(x, y) = g (f (x, y)) is continuous at(x0, y0).

Remark. We’ve already seen a special case of this theorem earliertoday?

Everything we discussed today has analogues in three or moredimensions. Consult the text.