Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 53



Work Step by Step

$$A=\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{e^x-1}\Big)$$ We unfortunately cannot do the replacement now since it would lead to the form $\infty-\infty$, which is a dead-end. $$A=\lim_{x\to0^+}\frac{e^x-1-x}{x(e^x-1)}$$ $\lim_{x\to0^+}(e^x-1-x)=(e^0-1-0)=1-1=0$ and $\lim_{x\to0^+}[x(e^x-1)]=0(e^0-1)=0.$ We have an indeterminate form of $\frac{0}{0}$, which can be dealt with by L'Hospital's Rule: $$A=\lim_{x\to0^+}\frac{(e^x-1-x)'}{[x(e^x-1)']}$$ $$A=\lim_{x\to0^+}\frac{e^x-1}{(e^x-1)+x(e^x-0)}$$ $$A=\lim_{x\to0^+}\frac{e^x-1}{e^x-1+xe^x}.$$ $\lim_{x\to0^+}(e^x-1)=e^0-1=1-1=0$ and $\lim_{x\to0^+}(e^x-1+xe^x)=e^0-1+0\times e^0=1-1+0=0.$ Another indeterminate form of $\frac{0}{0}$. We continue to take advantage of L'Hospital's Rule: $$A=\lim_{x\to0^+}\frac{(e^x-1)'}{(e^x-1+xe^x)'}$$ $$A=\lim_{x\to0^+}\frac{e^x}{e^x-0+(e^x+xe^x)}$$ $$A=\lim_{x\to0^+}\frac{e^x}{2e^x+xe^x}$$ $$A=\frac{e^0}{2e^0+0\times e^0}$$ $$A=\frac{1}{2\times1+0}$$ $$A=\frac{1}{2}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.