Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 37

Answer

$$\lim\limits_{x\to\infty}\frac{1-e^x}{1+2e^x}=\frac{-1}{2}$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{1-e^x}{1+2e^{x}}$$$$=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $e^x$, we have - In the numerator: $X=\frac{1-e^x}{e^x}=\frac{1}{e^x}-1$ - In the denominator: $Y=\frac{1+2e^x}{e^x}=\frac{1}{e^x}+2$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+2}$$ $$A=\frac{\lim\limits_{x\to\infty}(\frac{1}{e^x})-1}{\lim\limits_{x\to\infty}(\frac{1}{e^x})+2}$$ As $x\to\infty$, $e^x\to\infty$. So, $\frac{1}{e^x}\to0$. Which means, $\lim\limits_{x\to\infty}(\frac{1}{e^x})=0$ $$A=\frac{0-1}{0+2}=\frac{-1}{2}$$
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