James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

SectION 14.2 Limits and Continuity 907
Figure 7 shows the graph of the function
in Example 3. Notice the ridge
above the parabola x − y 2 .
limit is 0, for if we now let sx, yd l s0, 0d along the parabola x − y 2 , we have
f sx, yd − f sy 2 , yd − y 2 ? y 2
sy 2 d 2 1 y 4 − y 4
2y 4 − 1 2
0.5
z 0
_0.5
2
FIGURE 7
7et140207
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MasterID: 01565
0
x
_2
_2
0
2 y
so f sx, yd l 1 2
as sx, yd l s0, 0d along x − y
2
Since different paths lead to different limiting values, the given limit does not exist.
Now let’s look at limits that do exist. Just as for functions of one variable, the calculation
of limits for functions of two variables can be greatly simplified by the use of properties
of limits. The Limit Laws listed in Section 2.3 can be extended to functions of two
variables: the limit of a sum is the sum of the limits, the limit of a product is the product
of the limits, and so on. In particular, the following equations are true.
2 lim
sx, yd l sa, bd x − a
The Squeeze Theorem also holds.
lim y − b
sx, yd l sa, bd
lim c − c
sx, yd l sa, bd
■
EXAMPLE 4 Find
lim
sx, yd l s0, 0d
3x 2 y
x 2 2
if it exists.
1 y
SOLUTION As in Example 3, we could show that the limit along any line through the
origin is 0. This doesn’t prove that the given limit is 0, but the limits along the parabolas
y − x 2 and x − y 2 also turn out to be 0, so we begin to suspect that the limit does
exist and is equal to 0.
Let « . 0. We want to find . 0 such that
if 0 , sx 2 1 y 2 , then Z
3x 2 y
x 2 1 y 2 2 0 Z , «
that is, if 0 , sx 2 1 y 2 , then 3x 2 | y |
x 2 1 y 2 , «
But x 2 < x 2 1 y 2 since y 2 > 0, so x 2 ysx 2 1 y 2 d < 1 and therefore
3
3x 2 | y |
x 2 1 y 2 < 3 | y | − 3sy 2 < 3sx 2 1 y 2
Another way to do Example 4 is to
use the Squeeze Theorem instead of
Definition 1. From (2) it follows that
lim 3 | y | − 0
sx, yd l s0, 0d
and so the first inequality in (3) shows
that the given limit is 0.
Thus if we choose − «y3 and let 0 , sx 2 1 y 2 , , then
Hence, by Definition 1,
Z
3x 2 y
x 2 1 y 2 2 0 Z < 3sx 2 1 y 2
lim
sx, yd l s0, 0d
, 3 − 3S « 3D − «
3x 2 y
x 2 1 y 2 − 0 ■
Continuity
Recall that evaluating limits of continuous functions of a single variable is easy. It can
be accomplished by direct substitution because the defining property of a continuous
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