mathop {lim }limits_{x to 0} frac{{cos (sin x | vedclass

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Question

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}}$Using the trigonometric identity $\cos 	heta

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}}$Using the trigonometric identity $\cos 	heta - 1 = -2 \sin^2(	heta/2)$,we substitute $	heta = \sin x$:$= \mathop {\lim }\limits_{x 	o 0} \frac{{ - 2 \sin^2(\frac{\sin x}{2})}}{{{x^2}}}$$= -2 \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin(\frac{\sin x}{2})}{\frac{\sin x}{2}} \cdot \frac{\sin x}{2x} ight)^2$Since $\mathop {\lim }\limits_{u 	o 0} \frac{\sin u}{u} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we get:$= -2 \cdot (1)^2 \cdot (1/2)^2 = -2 \cdot \frac{1}{4} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}} = $

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