Page 145 - Calculus Demystified

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CHAPTER 5
Since this last limit is −2, we conclude that Indeterminate Forms

2
4
2
lim x − (x + 4x + 5) 1/2 =−2.
x→+∞
EXAMPLE 5.13
Evaluate

−x −3x 4 1/3
lim e − (e − x ) .
x→−∞
SOLUTION
First rewrite the limit as

4 3x 1/3

1 − (1 − x e )
4 3x 1/3
lim e −x 1 − (1 − x e ) = lim .
x→−∞ x→−∞ e x
Notice that both the numerator and denominator tend to zero (here we use the
4 3x
result analogous to Example 5.7 that x e → 0). So our new expression is
indeterminate of the form 0/0. l’Hôpital’s Rule applies and our limit equals
4 3x −2/3
3
4
3x
−(1/3)(1 − x e ) · (−4x · e 3x − x · 3e )
lim
x→−∞ e x
2x
4
4 3x −2/3
3
= lim (1/3)(1 − x e ) (4x · e 2x + 3x · e ).
x→−∞
2x
4
4
3x 3
Just as in Example 5.7, x ·e x ·e , and x ·e 2x all tend to zero. We conclude
that our limit equals 0.
√ √
You Try It: Evaluate lim x→+∞ [ x + 1 − x].
5.3 Improper Integrals: A First Look
5.3.1 INTRODUCTION
The theory of the integral that we learned earlier enables us to integrate a continuous
function f(x) on a closed, bounded interval [a, b]. See Fig. 5.1. However, it is
frequently convenient to be able to integrate an unbounded function, or a function
defined on an unbounded interval. In this section and the next we learn to do so, and
we see some applications of this new technique. The basic idea is that the integral
of an unbounded function is the limit of integrals of bounded functions; likewise,
the integral of a function on an unbounded interval is the limit of the integral on
bounded intervals.