mathop {lim }limits_{x to 0} frac{{log cos x} | vedclass

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(A) Using $L'	ext{Hospital's rule}$ as the limit is of the form $\frac{0}{0}$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\log \cos x}}{x} = \mathop {\l

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(A) Using $L'	ext{Hospital's rule}$ as the limit is of the form $\frac{0}{0}$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(\log \cos x)}}{{\frac{d}{{dx}}(x)}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{1}$$= \mathop {\lim }\limits_{x 	o 0} (-	an x)$$= -	an(0) = 0$

$\mathop {\lim }\limits_{x \to 0} \frac{{\log \cos x}}{x} = $

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