mathop {lim }limits_{x to 0} frac{{sin (2 + x | vedclass

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(C) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$:Let $C = 2+x$ and $D =

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$2\cos 2$

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(C) Using the trigonometric identity $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} ight) \sin \left( \frac{C-D}{2} ight)$:Let $C = 2+x$ and $D = 2-x$.Then $\sin(2+x) - \sin(2-x) = 2 \cos \left( \frac{2+x+2-x}{2} ight) \sin \left( \frac{2+x-(2-x)}{2} ight) = 2 \cos(2) \sin(x)$.Substituting this into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{2 \cos 2 \sin x}{x} = 2 \cos 2 \cdot \mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x}$.Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$,the result is $2 \cos 2 \cdot 1 = 2 \cos 2$.

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

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