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: 27

Answer

The solution is $$\lim_{x\to0}\frac{e^x-\sin x-1}{x^4+8x^3+12x^2}=\frac{1}{24}.$$

Work Step by Step

We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0}\frac{e^x-\sin x-1}{x^4+8x^3+12x^2}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(e^x-\sin x-1)'}{(x^4+8x^3+12x^2)'}=\lim_{x\to0}\frac{e^x-\cos x}{4x^3+24x^2+24x}=\left[\frac{e^0-\cos 0}{4\cdot0^3+24\cdot0^2+24\cdot0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(e^x-\cos x)'}{(4x^3+24x^2+24x)'}=\lim_{x\to0}\frac{e^x+\sin x}{12x^2+48x+24}=\frac{e^0+\sin 0}{12\cdot0^2+48\cdot0+24}=\frac{1}{24}.$$
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