Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Problems Plus - Problems - Page 274: 25

Answer

$$ \lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}=-\sin a $$

Work Step by Step

$$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}= \\=& \lim _{x \rightarrow 0} \frac{\sin a \cos 2 x+\cos a \sin 2 x-2 \sin a \cos x-2 \cos a \sin x+\sin a}{x^{2}}= \\=& \lim _{x \rightarrow 0} \frac{\sin a(\cos 2 x-2 \cos x+1)+\cos a(\sin 2 x-2 \sin x)}{x^{2}} =\\=& \lim _{x \rightarrow 0} \frac{\sin a\left(2 \cos ^{2} x-1-2 \cos x+1\right)+\cos a(2 \sin x \cos x-2 \sin x)}{x^{2}} =\\ &=\lim _{x \rightarrow 0} \frac{\sin a(2 \cos x)(\cos x-1)+\cos a(2 \sin x)(\cos x-1)}{x^{2}}= \\ & =\lim _{x \rightarrow 0} \frac{2(\cos x-1)[\sin a \cos x+\cos a \sin x](\cos x+1)}{x^{2}(\cos x+1)} =\\ & =\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x[\sin (a+x)]}{x^{2}(\cos x+1)}= \\ &=-2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \cdot \frac{\sin (a+x)}{\cos x+1}=\\ & =-2(1)^{2} \frac{\sin (a+0)}{\cos 0+1}=\\ &=-\sin a \end{aligned} $$ So , $$ \lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}=-\sin a $$
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