lim _{x rightarrow 0}(sin x)^{2 tan x} is equ | vedclass

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(B) Let $y = \lim _{x ightarrow 0} (\sin x)^{2 	an x}$.Taking the natural logarithm on both sides:$\ln y = \lim _{x ightarrow 0} 2 	an x \ln(\si

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(B) Let $y = \lim _{x ightarrow 0} (\sin x)^{2 	an x}$.Taking the natural logarithm on both sides:$\ln y = \lim _{x ightarrow 0} 2 	an x \ln(\sin x)$.This is an indeterminate form of type $0 	imes (-\infty)$. We rewrite it as:$\ln y = 2 \lim _{x ightarrow 0} \frac{\ln(\sin x)}{\cot x}$.Applying $L'H\hat{o}pital's$ rule:$\ln y = 2 \lim _{x ightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\csc^2 x} = 2 \lim _{x ightarrow 0} \frac{\cos x / \sin x}{-1 / \sin^2 x} = 2 \lim _{x ightarrow 0} (-\cos x \sin x)$.$\ln y = 2 	imes (-1 	imes 0) = 0$.Since $\ln y = 0$,we have $y = e^0 = 1$.

$\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$ is equal to

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