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$\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$ is equal to
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Let $y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$
$\Rightarrow \log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 \tan x}$
$=2 \lim _{x \rightarrow 0} \tan x \log \sin x$
$=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x}$
$=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos}{-cosec^{2}x} x$
$=2 \lim _{x \rightarrow 0}(-\sin x \cos x)$
$=2 \times 0=0$
$\therefore \quad \log y=0 \Rightarrow y=e^{0}=1$
$\Rightarrow \log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 \tan x}$
$=2 \lim _{x \rightarrow 0} \tan x \log \sin x$
$=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x}$
$=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos}{-cosec^{2}x} x$
$=2 \lim _{x \rightarrow 0}(-\sin x \cos x)$
$=2 \times 0=0$
$\therefore \quad \log y=0 \Rightarrow y=e^{0}=1$
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