limx→0cos(sinx)-cosxx4 is equal to:

limx→0cos(sinx)-cosxx4;00 =limx→02·sinx+sinx2·sinx-sinx2x4 =limx→02sinx+sinx2x+sinx2sinx-sinx2x-sinx2x+sinx2x4x-sinx2=limx→0x2-sin2x2x4:00 =limx→02x-2sinxcosx2·4·x3 by L-Hospital rule =limx→02x-sin2x8x3; =limx→02-2cos(2x)8(3)x2 =limx→02(1-cos(2x))244x2=424=16

limx→0cos(sinx)-cosxx4 is equal to:

limx→0cos(sinx)-cosxx4;00 =limx→02·sinx+sinx2·sinx-sinx2x4 =limx→02sinx+sinx2x+sinx2sinx-sinx2x-sinx2x+sinx2x4x-sinx2=limx→0x2-sin2x2x4:00 =limx→02x-2sinxcosx2·4·x3 by L-Hospital rule =limx→02x-sin2x8x3; =limx→02-2cos(2x)8(3)x2 =limx→02(1-cos(2x))244x2=424=16

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

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$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{12}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}}; (\frac{0}{0}) = \lim_{x \to 0} (\frac{2 \cdot \sin (\frac{x + \sin x}{2}) \cdot \sin (\frac{x - \sin x}{2})}{x^{4}})$

$= \lim_{x \to 0} 2 (\frac{\sin (\frac{x + \sin x}{2})}{(\frac{x + \sin x}{2})}) (\frac{\sin (\frac{x - \sin x}{2})}{(\frac{x - \sin x}{2})}) (\frac{\frac{x + \sin x}{2})}{x^{4}} (\frac{x - \sin x}{2})$

$= \lim_{x \to 0} (\frac{x^{2} - \sin^{2} x}{2 x^{4}}) : (\frac{0}{0})$

$= \lim_{x \to 0} \frac{2 x - 2 \sin x \cos x}{2 \cdot 4 \cdot x^{3}} (b y L - H o s p i t a l r u l e) = \lim_{x \to 0} \frac{2 x - \sin 2 x}{8 x^{3}}; = \lim_{x \to 0} \frac{2 - 2 \cos (2 x)}{8 (3) x^{2}} = \lim_{x \to 0} \frac{2 (1 - \cos (2 x))}{24 (4 x^{2})} = \frac{4}{24} = \frac{1}{6}$