mathop {lim }limits_{x to 0} frac{{ln (cos x) | vedclass

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(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$.Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with res

Vedclass · Accepted Answer

$-\frac{1}{2}$

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(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$.Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\ln (\cos x)}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{{2x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{-	an x}}{{2x}}$Applying $L'Hospital's$ rule again:$= \mathop {\lim }\limits_{x 	o 0} \frac{{-\sec^2 x}}{2}$$= \frac{{-1^2}}{2} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$ is equal to

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