University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Additional and Advanced Exercises - Page 114: 25


$$\lim_{x\to0}\frac{\sin(1-\cos x)}{x}=0$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\sin(1-\cos x)}{x}$$ $$A=\lim_{x\to0}\frac{\sin(1-\cos x)}{1-\cos x}\times\lim_{x\to0}\frac{1-\cos x}{x}$$ - For all $x\ne0$, $(1-\cos x)\ne0$. So $\lim_{x\to0}\frac{\sin(1-\cos x)}{1-\cos x}=1$ $$A=1\times\lim_{x\to0}\frac{1-\cos x}{x}$$ $$A=\lim_{x\to0}\frac{1-\cos x}{x}$$ Multiply both numerator and denominator by $1+\cos x$: $$A=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{1-\cos^2 x}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin^2x}{x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin x}{x}\times\lim_{x\to0}\frac{\sin x}{1+\cos x}$$ $$A=1\times\Big(\frac{\sin0}{1+\cos0}\Big)$$ $$A=\frac{0}{1+1}=0$$
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