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: 32

Answer

$\lim\limits_{x\to\infty}(e^{-x}+2\cos{3x})$ does not exist.

Work Step by Step

$$\lim\limits_{x\to\infty}(e^{-x}+2\cos3x)$$$$=\lim\limits_{x\to\infty}e^{-x}+2\lim\limits_{x\to\infty}\cos3x$$$$=\lim\limits_{x\to\infty}(\frac{1}{e^x})+2\lim\limits_{x\to\infty}\cos3x$$$$=\frac{1}{\lim\limits_{x\to\infty}e^x}+2\lim\limits_{x\to\infty}\cos3x$$$$=A+B$$ - Consider A: As $x\to\infty$, $e^x$ approaches $\infty$. So, $\lim\limits_{x\to\infty}e^x=\infty$ Therefore, $A=\frac{1}{\lim\limits_{x\to\infty}e^x}=0$ - Consider B: We know that $\cos3x$ oscillates constantly between $1$ and $-1$ as $x$ increases, so it does not reach any definite number. In other words, $\lim\limits_{x\to\infty}\cos{3x}$ does not exist. Therefore, $A+B$ does not exist. Overall, $\lim\limits_{x\to\infty}(e^{-x}+2\cos{3x})$ does not exist.
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