_ x 0 ( x)-1 x^2 = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-1}{x^2}=$

A
1
B
-1
C
$\frac{1}{2}$
✓
$-\frac{1}{2}$

✓

Answer

Correct option: D.

$-\frac{1}{2}$

(D)
$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-1}{x^2}=\lim _{x \rightarrow 0} \frac{-2 \sin ^2\left(\frac{\sin x}{2}\right)}{x^2}$
$=-2 \lim _{x \rightarrow 0} \frac{\sin ^2\left(\frac{\sin x}{2}\right)}{\frac{\sin ^2 x}{4}} \times \frac{\sin ^2 x}{4 x^2}$
$=-\frac{1}{2} \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{\sin x}{2}\right)}{\frac{\sin x}{2}}\right]^2 \times\left(\frac{\sin x}{x}\right)^2$
$=-\frac{1}{2}(1)^2(1)^2=\frac{-1}{2}$

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