lim _{x rightarrow 0} frac{1-cos (1-cos x)}{s | vedclass

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Question

(D) Given that,$\lim _{x ightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$ $= \lim _{x}$ ${ightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} 	i

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(D) Given that,$\lim _{x ightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$ $= \lim _{x}$ ${ightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} 	imes \left(\frac{1-\cos x}{x^2}ight)^2 	imes \left(\frac{x}{\sin x}ight)^4$ As $x ightarrow 0$,$(1-\cos x) ightarrow 0$. Using the standard limit $\lim _{	heta ightarrow 0} \frac{1-\cos 	heta}{	heta^2} = \frac{1}{2}$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$: $= \frac{1}{2} 	imes \left(\frac{1}{2}ight)^2 	imes (1)^4$ $= \frac{1}{2} 	imes \frac{1}{4} 	imes 1 = \frac{1}{8}$

$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

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