mathop {lim }limits_{x to 0} x log (sin x) = | vedclass

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(A) We need to evaluate $L = \mathop {\lim }\limits_{x 	o 0} x \log (\sin x)$.As $x 	o 0^+$,$\sin x 	o 0^+$,so $\log(\sin x) 	o -\infty$.This is a

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(A) We need to evaluate $L = \mathop {\lim }\limits_{x 	o 0} x \log (\sin x)$.As $x 	o 0^+$,$\sin x 	o 0^+$,so $\log(\sin x) 	o -\infty$.This is an indeterminate form of type $0 	imes (-\infty)$.We can rewrite the expression as $L = \mathop {\lim }\limits_{x 	o 0} \frac{\log(\sin x)}{1/x}$.Applying $L$'$H$ôpital's rule:$L = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{1}{\sin x} \cdot \cos x}{-1/x^2} = \mathop {\lim }\limits_{x 	o 0} -x^2 \cot x$.$L = \mathop {\lim }\limits_{x 	o 0} -x \cdot (x \cot x) = \mathop {\lim }\limits_{x 	o 0} -x \cdot \frac{x}{	an x}$.Since $\mathop {\lim }\limits_{x 	o 0} \frac{x}{	an x} = 1$,we have $L = 0 \cdot 1 = 0$.

$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

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