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: 41

Answer

$$\lim_{x\to0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}=\frac{1}{24}$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}$$ Here $\lim_{x\to0}(\cos x-1+\frac{1}{2}x^2)=\cos0-1+\frac{1}{2}0^2=1-1+0=0$ and $\lim_{x\to0}(x^4)=0^4=0.$ This limit is in an indeterminate form of $\frac{0}{0}$, a situation which can be helped by the use of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-\sin x-0+\frac{1}{2}\times2x}{4x^3}$$ $$A=\lim_{x\to0}\frac{-\sin x+x}{4x^3}$$ $\lim_{x\to0}(\sin x+x)=\sin0+0=0$ and $\lim_{x\to0}(4x^3)=4\times0^3=0.$ Again we have an indeterminate form of $\frac{0}{0}$. We continue to use L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-\cos x+1}{12x^2}$$ $\lim_{x\to0}(-\cos x+1)=-\cos0+1=-1+1=0$ and $\lim_{x\to0}(12x^2)=12\times0^2=0.$ Indeterminate form of $\frac{0}{0}$. We use L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-(-\sin x)+0}{24x}$$ $$A=\lim_{x\to0}\frac{\sin x}{24x}$$ $\lim_{x\to0}\sin x=\sin 0=0$ and $\lim_{x\to0}(24x)=24\times0=0.$ Sometimes it just goes on and on like this. Indeterminate form of $\frac{0}{0}$ for another application of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{\cos x}{24}$$ $$A=\frac{\cos0}{24}$$ $$A=\frac{1}{24}$$
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