limx→0 cos⁡(sin⁡x)−1x2= - Infinity Learn by Sri Chaitanya

A
1
B
$- 1$
C
$1 / 2$
D
$- 1 / 2$

Solution:

We have,

$\begin{array}{l} lim_{x \to 0} \frac{\cos (\sin x) - 1}{x^{2}} \\ = - lim_{x \to 0} \frac{1 - \cos (\sin x)}{\sin^{2} x} \times \frac{\sin^{2} x}{x^{2}} \end{array}$

$= - \frac{1}{2} \times 1 [∵ lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}]$

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