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: 36

Answer

$$\lim\limits_{x\to\infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}=1$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$$$$=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $e^{3x}$, we have - In the numerator: $X=\frac{e^{3x}-e^{-3x}}{e^{3x}}=1-e^{-6x}$ - In the denominator: $Y=\frac{e^{3x}+e^{-3x}}{e^{3x}}=1+e^{-6x}$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{1-e^{-6x}}{1+e^{-6x}}$$ $$A=\frac{1-\lim\limits_{x\to\infty}(e^{-6x})}{1+\lim\limits_{x\to\infty}(e^{-6x})}$$ $$A=\frac{1-\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})}{1+\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})}$$ As $x\to\infty$, $e^{6x}$ approaches $\infty$. So, $\frac{1}{e^{6x}}$ approaches $0$. Therefore, $\lim\limits_{x\to\infty}(\frac{1}{e^{6x}})=0$ So, $$A=\frac{1-0}{1+0}=1$$
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