Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 18

Answer

$$\lim\limits_{x\to\infty}e^{x-x^2}=0$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}e^{x-x^2}=\lim\limits_{x\to\infty}e^{x(1-x)}$$ As $x\to\infty$, $(1-x)$ approaches $-\infty$. That means $x(1-x)$ approaches $-\infty$. Therefore, $$A=e^{-\infty}=\frac{1}{e^{\infty}}=0$$
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