_ x 0 x x = - Vidyadip

MCQ

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

✓
$0$
B
$1$
C
$\infty $
D
None of these

✓

Answer

Correct option: A.

$0$

a
(a) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{\log \,\left[ {1 - 2{{\sin }^2}\frac{x}{2}} \right]}}{x}$

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \,\left[ {2\,{{\sin }^2}\frac{x}{2} + {{\left( {\frac{{2\,{{\sin }^2}\frac{x}{2}}}{2}} \right)}^2} + ......} \right]}}{x} = 0$

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \tan x}}{1} = 0.$

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