limx→0 cos⁡(sin⁡x)−cos⁡xx4=_____________

cos⁡C−cos⁡D=2sin⁡C+D2⋅sin⁡D−C2limx→0 cos⁡(sin⁡x)−cos⁡xx4==limx→0 2sin⁡(x+sin⁡x)2⋅sin⁡x−sin⁡x2x4=limx→0 2sin⁡x+sin⁡x2x+sin⁡x2sin⁡x−sin⁡x2x−sin⁡x2x+sin⁡x2x−sin⁡x2x4=limx→0 x2−sin2⁡x2x4L′H Rule =limx→0 2x−sin⁡2x8x3L′H Rule =limx→0 2−2cos⁡2x24x2L′H Rule =limx→0 +4sin⁡2x48x=limx→0 8cos⁡2x48=16

limx→0 cos⁡(sin⁡x)−cos⁡xx4=_____________

cos⁡C−cos⁡D=2sin⁡C+D2⋅sin⁡D−C2limx→0 cos⁡(sin⁡x)−cos⁡xx4==limx→0 2sin⁡(x+sin⁡x)2⋅sin⁡x−sin⁡x2x4=limx→0 2sin⁡x+sin⁡x2x+sin⁡x2sin⁡x−sin⁡x2x−sin⁡x2x+sin⁡x2x−sin⁡x2x4=limx→0 x2−sin2⁡x2x4L′H Rule =limx→0 2x−sin⁡2x8x3L′H Rule =limx→0 2−2cos⁡2x24x2L′H Rule =limx→0 +4sin⁡2x48x=limx→0 8cos⁡2x48=16

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

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$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{4}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \cos C - \cos D = 2 \sin \frac{C + D}{2} \cdot \sin \frac{D - C}{2} \\ lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}} = \\ = lim_{x \to 0} 2 \frac{\frac{\sin (x + \sin x)}{2} \cdot \sin (\frac{x - \sin x}{2})}{x^{4}} \\ = lim_{x \to 0} 2 (\frac{\sin (\frac{x + \sin x}{2})}{(\frac{x + \sin x}{2})}) (\frac{\sin (\frac{x - \sin x}{2})}{\frac{x - \sin x}{2}}) (\frac{(\frac{x + \sin x}{2}) (\frac{x - \sin x}{2})}{x^{4}} \\ = lim_{x \to 0} \frac{x^{2} - \sin^{2} x}{2 x^{4}} L^{'} H Rule = lim_{x \to 0} \frac{2 x - \sin 2 x}{8 x^{3}} L^{'} H Rule \\ = lim_{x \to 0} \frac{2 - 2 \cos 2 x}{24 x^{2}} L^{'} H Rule = lim_{x \to 0} \frac{+ 4 \sin 2 x}{48 x} = lim_{x \to 0} \frac{8 \cos 2 x}{48} = \frac{1}{6} \end{array}$