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

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

Class 11 Mathematics Limits L'Hospital's rule and Limit of Indeterminate Form

Easy

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

A
$0$
B
$1$
C
$\infty$
D
$\text{None of these}$

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Class 11

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Limits

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L'Hospital's rule and Limit of Indeterminate Form

If $f(9)=9$ and $f^{\prime}(9)=4$,then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$

MediumAP EAMCET 2017

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The value of $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{2/x}}$ where $(a, b, c > 0)$ is:

Difficult

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$\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ is equal to

DifficultJEE MAIN 2022

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If $f(5)=7$ and $f'(5)=7$, then $\lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$ is given by

EasyWBJEE 2010

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

Similar Questions

If $f(9)=9$ and $f^{\prime}(9)=4$,then $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

The value of $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)^{2/x}}$ where $(a, b, c > 0)$ is:

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ is equal to

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