lim _{x rightarrow 0} frac{8}{sin ^8 x} left{ | vedclass

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Question

(B) Let the expression be $L = \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\c

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) Let the expression be $L = \lim _{x ightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}ight)-\cos \left(\frac{x^2}{4}ight)+\cos \left(\frac{x^2}{2}ight) \cos \left(\frac{x^2}{4}ight)ight\}$.Factoring the expression inside the curly brackets:$1-\cos \left(\frac{x^2}{2}ight)-\cos \left(\frac{x^2}{4}ight)+\cos \left(\frac{x^2}{2}ight) \cos \left(\frac{x^2}{4}ight) = \left(1-\cos \left(\frac{x^2}{2}ight)ight) - \cos \left(\frac{x^2}{4}ight) \left(1-\cos \left(\frac{x^2}{2}ight)ight) = \left(1-\cos \left(\frac{x^2}{2}ight)ight) \left(1-\cos \left(\frac{x^2}{4}ight)ight)$.Using the identity $1-\cos 	heta = 2 \sin ^2 \left(\frac{	heta}{2}ight)$:$L = \lim _{x ightarrow 0} \frac{8}{\sin ^8 x} \left(2 \sin ^2 \left(\frac{x^2}{4}ight)ight) \left(2 \sin ^2 \left(\frac{x^2}{8}ight)ight) = \lim _{x ightarrow 0} \frac{32 \sin ^2 \left(\frac{x^2}{4}ight) \sin ^2 \left(\frac{x^2}{8}ight)}{\sin ^8 x}$.Using the limit $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$:$L = 32 \lim _{x ightarrow 0} \left[ \frac{\sin ^2 \left(\frac{x^2}{4}ight)}{\left(\frac{x^2}{4}ight)^2} \cdot \left(\frac{x^2}{4}ight)^2 \cdot \frac{\sin ^2 \left(\frac{x^2}{8}ight)}{\left(\frac{x^2}{8}ight)^2} \cdot \left(\frac{x^2}{8}ight)^2 \cdot \frac{1}{\left(\frac{\sin x}{x}ight)^8 \cdot x^8} ight]$.$L = 32 \cdot 1 \cdot \frac{x^4}{16} \cdot 1 \cdot \frac{x^4}{64} \cdot \frac{1}{x^8} = 32 \cdot \frac{1}{16 \cdot 64} = \frac{32}{1024} = \frac{1}{32}$.

$\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\} =$

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