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: 40

Answer

$$\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x-\sin x}=2$$

Work Step by Step

$$A=\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x-\sin x}$$ $\lim_{x\to0}(e^x-e^{-x}-2x)=e^0-e^{-0}-2\times0=1-1-0=0$ and $\lim_{x\to0}(x-\sin x)=0-\sin0=0,$ which is the indeterminate form $\frac{0}{0}$. L'Hospital's Rule can be used: $$A=\lim_{x\to0}\frac{(e^x-e^{-x}-2x)'}{(x-\sin x)'}$$ $$A=\lim_{x\to0}\frac{e^x-e^{-x}(-x)'-2}{1-\cos x}$$ $$A=\lim_{x\to0}\frac{e^x-e^{-x}(-1)-2}{1-\cos x}$$ $$A=\lim_{x\to0}\frac{e^x+e^{-x}-2}{1-\cos x}$$ $\lim_{x\to0}(e^x+e^{-x}-2)=e^0+e^{-0}-2=1+1-2=0$ and $\lim_{x\to0}(1-\cos x)=1-\cos0=1-1=0.$ Another indeterminate form: $\frac{0}{0}$. We use L'Hospital's Rule one more time: $$A=\lim_{x\to0}\frac{(e^x+e^{-x}-2)'}{(1-\cos x)'}$$ $$A=\lim_{x\to0}\frac{e^x+e^{-x}(-x)'}{0-(-\sin x)}$$ $$A=\lim_{x\to0}\frac{e^x-e^{-x}}{\sin x}$$ $\lim_{x\to0}(e^x-e^{-x})=e^0-e^{-0}=1-1=0$ and $\lim_{x\to0}\sin x=\sin0=0.$ Apparently, this indeterminate form of $\frac{0}{0}$ still has one last shot. We still can use L'Hospital's Rule: $$A=\lim_{x\to0}\frac{(e^x-e^{-x})'}{(\sin x)'}$$ $$A=\lim_{x\to0}\frac{e^x+e^{-x}}{\cos x}\hspace{1cm}(\text{This is finally over.})$$ $$A=\frac{e^0+e^{-0}}{\cos 0}$$ $$A=\frac{1+1}{1}$$ $$A=2$$
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.