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: 42

Answer

$$\lim\limits_{x\to\infty}[\ln(2+x)-\ln(1+x)]=0$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}[\ln(2+x)-\ln(1+x)]$$$$A=\lim\limits_{x\to\infty}\Bigg[\ln\Big(\frac{2+x}{1+x}\Big)\Bigg]$$$$A=\ln\Bigg[\lim\limits_{x\to\infty}\Big(\frac{2+x}{1+x}\Big)\Bigg]$$ Divide both numerator and denominator by $x$, we have - The numerator: $\frac{2+x}{x}=\frac{2}{x}+1$ - The denominator: $\frac{1+x}{x}=\frac{1}{x}+1$ Therefore, $$A=\ln\Bigg[\lim\limits_{x\to\infty}\Big(\frac{\frac{2}{x}+1}{\frac{1}{x}+1}\Big)\Bigg]$$ $$A=\ln\Bigg[\frac{2\times0+1}{0+1}\Bigg]$$ $$A=\ln1=0$$
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