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: 41

Answer

$$ \lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} =\frac{1}{24} $$

Work Step by Step

$$ \lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} $$ Since $$ \lim _{x \rightarrow 0} ( \cos x-1+\frac{1}{2} x^{2})=0 $$ and $$ \lim _{x \rightarrow 0} ( x^{4})=0 $$ the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule: $$ \begin{aligned} \lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} & \stackrel{H}{=} \lim _{x \rightarrow 0} \frac{-\sin x+x}{4 x^{3}} \rightarrow \frac{0}{0} \\ &\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{-\cos x+1}{12 x^{2}} \rightarrow \frac{0}{0} \\ & \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\sin x}{24 x} \rightarrow \frac{0}{0} \\ & \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{\cos x}{24} \\ &=\frac{1}{24} \end{aligned} $$
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