limx→0cos(sinx)−cosxx4=

limx→0cos(sinx)−cosxx4=limx→0−2x4sin(sinx+x2)sin(sinx−x2)=limx→0−2sin(sinx+x2)(sinx+x2).sin(sinx−x2)sin(sinx−x2)×(sinx+xx)(sinx−xx3)14=−24×1×1×limx→0(sinxx+1)×limx→0(x−x33!+x55!.....xx3)=−12×(1+1)×(−16)=16

limx→0cos(sinx)−cosxx4=

limx→0cos(sinx)−cosxx4=limx→0−2x4sin(sinx+x2)sin(sinx−x2)=limx→0−2sin(sinx+x2)(sinx+x2).sin(sinx−x2)sin(sinx−x2)×(sinx+xx)(sinx−xx3)14=−24×1×1×limx→0(sinxx+1)×limx→0(x−x33!+x55!.....xx3)=−12×(1+1)×(−16)=16

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$\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}} =$

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$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

$\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}}$ = $\lim_{x \to 0} - \frac{2}{x^{4}} \sin (\frac{\sin x + x}{2}) \sin (\frac{\sin x - x}{2})$ = $\lim_{x \to 0} - 2 \frac{\sin (\frac{\sin x + x}{2})}{(\frac{\sin x + x}{2})} . \frac{\sin (\frac{\sin x - x}{2})}{\sin (\frac{\sin x - x}{2})} \times (\frac{\sin x + x}{x}) (\frac{\sin x - x}{x^{3}}) \frac{1}{4}$
$= \frac{- 2}{4} \times 1 \times 1 \times \lim_{x \to 0} (\frac{\sin x}{x} + 1) \times \lim_{x \to 0} (\frac{x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} ..... x}{x^{3}})$ $= \frac{- 1}{2} \times (1 + 1) \times (- \frac{1}{6}) = \frac{1}{6}$