Evaluate the limit: lim _{x} {rightarrow 0}le | vedclass

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Question

(C) Let the given expression be $L = \lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3}

Vedclass · Accepted Answer

$\frac{1}{24}$

Vedclass · Answer

(C) Let the given expression be $L = \lim _{x}$ ${ightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}ight)ight)$.We can factor the expression inside the limit as:$L = \lim _{x ightarrow 0} \frac{4!}{x^8} (1 - \cos \frac{x^2}{3})(1 - \cos \frac{x^2}{4})$.Using the identity $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$,we get:$1 - \cos \frac{x^2}{3} = 2 \sin^2(\frac{x^2}{6})$ and $1 - \cos \frac{x^2}{4} = 2 \sin^2(\frac{x^2}{8})$.Substituting these into the limit:$L = 24 	imes \lim _{x ightarrow 0} \frac{1}{x^8} 	imes 2 \sin^2(\frac{x^2}{6}) 	imes 2 \sin^2(\frac{x^2}{8})$.$L = 24 	imes 4 	imes \lim _{x}$ ${ightarrow 0} \left(\frac{\sin(\frac{x^2}{6})}{\frac{x^2}{6}}ight)^2 	imes (\frac{x^2}{6})^2 	imes \left(\frac{\sin(\frac{x^2}{8})}{\frac{x^2}{8}}ight)^2 	imes (\frac{x^2}{8})^2 	imes \frac{1}{x^8}$.Since $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have:$L = 96 	imes 1^2 	imes \frac{x^4}{36} 	imes 1^2 	imes \frac{x^4}{64} 	imes \frac{1}{x^8}$.$L = 96 	imes \frac{1}{36 	imes 64} = \frac{96}{2304} = \frac{1}{24}$.

Evaluate the limit: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$

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