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: 54

Answer

$$\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{\tan^{-1}x}\Big)=0$$

Work Step by Step

$$A=\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{\tan^{-1}x}\Big)$$ $$A=\lim_{x\to0^+}\frac{\tan^{-1}x-x}{x\tan^{-1}x}$$ $\lim_{x\to0^+}(\tan^{-1}x-x)=\tan^{-1}0-0=0$ and $\lim_{x\to0^+}(x\tan^{-1}x)=0\times\tan^{-1}0=0.$ For this being an indeterminate form of $\frac{0}{0}$, L'Hospital's Rule can be used: $$A=\lim_{x\to0^+}\frac{(\tan^{-1}x-x)'}{(x\tan^{-1}x)'}$$ (Don't forget that $(\tan^{-1}x)'=\frac{1}{1+x^2}$) $$A=\lim_{x\to0^+}\frac{\frac{1}{1+x^2}-1}{\tan^{-1}x+x\times\frac{1}{1+x^2}}$$ $$A=\lim_{x\to0^+}\frac{\frac{1-(1+x^2)}{1+x^2}}{\frac{\tan^{-1}x(1+x^2)+x}{1+x^2}}$$ $$A=\lim_{x\to0^+}\frac{\frac{-x^2}{1+x^2}}{\frac{\tan^{-1}x(1+x^2)+x}{1+x^2}}$$ $$A=\lim_{x\to0^+}\frac{-x^2}{\tan^{-1}x(1+x^2)+x}.$$ $\lim_{x\to0^+}(-x^2)=-0^2=0$ and $\lim_{x\to0^+}(\tan^{-1}x(1+x^2)+x)=\tan^{-1}0(1+0^2)+0=0\times1+0=0.$ Another indeterminate form of $\frac{0}{0}$. We use L'Hospital's Rule one more time: $$A=\lim_{x\to0^+}\frac{(-x^2)'}{[\tan^{-1}x(1+x^2)+x]'}$$ $$A=\lim_{x\to0^+}\frac{-2x}{\frac{1}{1+x^2}(1+x^2)+\tan^{-1}x(2x)+1}$$ $$A=\lim_{x\to0^+}\frac{-2x}{1+\tan^{-1}x(2x)+1}$$ $$A=\lim_{x\to0^+}\frac{-2x}{2x\tan^{-1}x+2}$$ $$A=\frac{-2\times0}{2\times0\times\tan^{-1}0+2}$$ $$A=\frac{0}{0+2}=0$$
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