Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 312: 64


$\lim\limits_{x \to \infty}x^{e^{-x}} = 1$

Work Step by Step

$\lim\limits_{x \to \infty}x^{e^{-x}}=\lim\limits_{x \to \infty}(e^{ln~x})^{e^{-x}}=\lim\limits_{x \to \infty}e^{ln~x/e^x}$ $\lim\limits_{x \to \infty}\frac{ln~x}{e^x} = \lim\limits_{x \to \infty}\frac{(1/x)}{e^x} = 0$ Therefore: $\lim\limits_{x \to \infty}e^{ln~x/e^x} = e^0 = 1$
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