## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 25

#### Answer

$1$

#### Work Step by Step

Find the limit$\lim\limits_{x \to \infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$. Thus, $\lim\limits_{x \to \infty}\frac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}=\lim\limits_{x \to \infty}\frac{e^{3x}(1-e^{-6x})}{e^{3x}(1+e^{-6x})}$ $=\frac{1-\lim\limits_{x \to \infty }e^{-6x}}{1+\lim\limits_{x \to \infty} e^{-6x}}$ $=\frac{1-0}{1+0}$ $=1$

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