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Calculus 2nd Edition Rogawski

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SECTION 15.2 Limits and Continuity in Several Variables 795<br />

z<br />

EXAMPLE 2 Evaluating Limits by Substitution Show that<br />

y<br />

f (x, y) = 3x + y<br />

x 2 + y 2 + 1<br />

is continuous (Figure 4). Then evaluate lim f (x, y).<br />

(x,y)→(1,2)<br />

Solution The function f (x, y) is continuous at all points (a, b) because it is a rational<br />

function whose denominator Q(x, y) = x 2 + y 2 + 1 is never zero. Therefore, we can<br />

x<br />

evaluate the limit by substitution:<br />

3x + y<br />

lim<br />

(x,y)→(1,2)<br />

FIGURE 4 Top view of the graph<br />

x 2 + y 2 + 1 = 3(1) + 2<br />

1 2 + 2 2 + 1 = 5 6<br />

f (x, y) =<br />

3x + y<br />

x 2 + y 2 + 1 . If f (x, y) is a product f (x, y) = h(x)g(y), where h(x) and g(y) are continuous,<br />

then the limit is a product of limits by the Product Law:<br />

( ) ( )<br />

lim f (x, y) = lim h(x)g(y) = lim h(x) lim g(y)<br />

(x,y)→(a,b) (x,y)→(a,b) x→a y→b<br />

EXAMPLE 3 Product Functions Evaluate lim<br />

(x,y)→(3,0)<br />

x3<br />

sin y<br />

y .<br />

Solution The limit is equal to a product of limits:<br />

lim<br />

(x,y)→(3,0)<br />

x3<br />

sin y<br />

y<br />

=<br />

(<br />

lim<br />

x→3 x3 )(<br />

lim<br />

y→0<br />

)<br />

sin y<br />

= (3 3 )(1) = 27<br />

y<br />

Composition is another important way to build functions. If f (x, y) is a function<br />

of two variables and G(u) a function of one variable, then the composite G ◦ f is the<br />

function G(f (x, y)). According to the next theorem, a composite of continuous functions<br />

is again continuous.<br />

THEOREM 2 A Composite of Continuous Functions Is Continuous If f (x, y) is continuous<br />

at (a, b) and G(u) is continuous at c = f (a, b), then the composite function<br />

G(f (x, y)) is continuous at (a, b).<br />

EXAMPLE 4 Write H (x, y) = e −x2 +2y as a composite function and evaluate<br />

lim H (x, y)<br />

(x,y)→(1,2)<br />

Solution We have H (x, y) = G ◦ f , where G(u) = e u and f (x, y) = −x 2 + 2y. Both<br />

f and G are continuous, so H is also continuous and<br />

lim H (x, y) = lim<br />

(x,y)→(1,2) (x,y)→(1,2) e−x2 +2y = e −(1)2 +2(2) = e 3<br />

We know that if a limit<br />

lim<br />

(x,y)→(a,b)<br />

f (x, y) exists and equals L, then f (x, y) tends to<br />

L as (x, y) approaches (a, b) along any path. In the next example, we prove that a limit<br />

does not exist by showing that f (x, y) approaches different limits along lines through the<br />

origin.

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