_ x 0 ( ^2 x) x^2 = - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}} = $

A
$( - 1,1)$
✓
$\pi $
C
$\pi /2$
D
$1$

✓

Answer

Correct option: B.

$\pi $

b
(b) Limit $ = \mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{{\cos (\pi {{\cos }^2}x).\pi .2\cos x( - \sin x)}}{{2x}}} \right)$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \pi \cos (\pi {\cos ^2}x).\cos x.\left( {\frac{{ - \sin x}}{x}} \right)$

$ = \pi ( - 1).1.( - 1) = \pi $.

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