Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 26

Answer

The solution is $$\lim_{x\to0}\frac{e^x-x-1}{5x^2}=\frac{1}{10}.$$

Work Step by Step

We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0}\frac{e^x-x-1}{5x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(e^x-x-1)'}{(5x^2)'}=\lim_{x\to0}\frac{e^x-1}{10x}=\left[\frac{e^0-1}{10\cdot1}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(e^x-1)'}{(10x)'}=\lim_{x\to0}\frac{e^x}{10}=\frac{e^0}{10}=\frac{1}{10}.$$
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