Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 53

Answer

$$ \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) =\frac{1}{2} $$

Work Step by Step

$$ \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) $$ First notice that $$ \lim _{x \rightarrow 0^{+}} \frac{1}{x} \,\,\, \rightarrow \infty $$ and $$ \lim _{x \rightarrow 0^{+}} \frac{1}{e^{x}-1} \,\,\, \rightarrow \infty , $$ so the limit is indeterminate and has the form $\infty -\infty $. Here we use a common denominator: $$\begin{aligned} \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) &=\lim _{x \rightarrow 0^{+}} \frac{e^{x}-1-x}{x\left(e^{x}-1\right)} \rightarrow \frac{0}{0} \\ & \stackrel{\text { H }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}-1}{x e^{x}+e^{x}-1} \rightarrow \frac{0}{0} \\ & \stackrel{\text { H }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}}{x e^{x}+e^{x}+e^{x}} \\ & =\frac{1}{0+1+1} \\ &=\frac{1}{2} \end{aligned} $$ Note that the use of l’Hospital’s Rule is justified because: $ e^{x}-1-x \rightarrow 0 \quad $ and $ \quad x\left(e^{x}-1\right) \rightarrow 0 $ as$ \rightarrow 0^{+}$ also, $ e^{x}-1 \rightarrow 0 \quad $ and $ \quad x e^{x}+e^{x}-1 \rightarrow 0 $ as$ \rightarrow 0^{+}$.
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