Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 317: 64

Answer

$1$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} \cr & {\text{Evaluating the limit when }}x \to \infty \cr & \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = {\infty ^0} \cr & {\text{This limit has the form }}{\infty ^0}{\text{ }} \cr & {x^{{e^{ - x}}}} = {e^{^{{e^{ - x}}}\ln x}} = {e^{\frac{{\ln x}}{{{e^x}}}}},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{{\ln x}}{{{e^x}}}}} = {e^{{{\mathop {\lim }\limits_{x \to \infty } }^{\frac{{\ln x}}{{{e^x}}}}}}} \cr & {\text{The first step is to evaluate }} \cr & L = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{e^x}}} = \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln x} \right]}}{{\frac{d}{{dx}}\left[ {{e^x}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{x}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{x{e^x}}} \cr & L = 0 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln x}} = {e^0} = {e^L} = 1 \cr & \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = {e^L} \cr & \mathop {\lim }\limits_{x \to \infty } {x^{{e^{ - x}}}} = 1 \cr} $$
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