mathop {lim }limits_{x to 0} frac{{1 - cos 2x | vedclass

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Question

(A) We know that $1 - \cos 2x = 2 \sin^2 x$.So,the limit becomes $\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2 x}{x}$.This can be rewritten as $\ma

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Vedclass · Answer

(A) We know that $1 - \cos 2x = 2 \sin^2 x$.So,the limit becomes $\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2 x}{x}$.This can be rewritten as $\mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \frac{\sin x}{x} \cdot \sin x ight)$.Using the standard limit $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we get $2 \cdot 1 \cdot \sin(0) = 2 \cdot 1 \cdot 0 = 0$.

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

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