mathop {lim }limits_{x to 0} frac{{1 - cos x} | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{1 - \cos x}}{{{{\sin }^2}x}}$.Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$ and $\sin^2 x = (2 \sin(\frac{x}{2}) \cos(\frac{x}{2}))^2 = 4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})$:$\mathop {\lim }\limits_{x 	o 0} \frac{{2 \sin^2(\frac{x}{2})}}{{4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})}} = \mathop {\lim }\limits_{x 	o 0} \frac{1}{2 \cos^2(\frac{x}{2})}$.As $x 	o 0$,$\cos(\frac{x}{2}) 	o \cos(0) = 1$.Therefore,the limit is $\frac{1}{2(1)^2} = \frac{1}{2}$.

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

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