If lim _{x rightarrow 0} frac{sin (2+x)-sin ( | vedclass

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(D) We use the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$.Given the limit $\lim _{x

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$2, 2$

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(D) We use the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$.Given the limit $\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x} = A \cos B$.Applying the identity with $C = 2+x$ and $D = 2-x$:$\lim _{x \rightarrow 0} \frac{2 \cos \left(\frac{2+x+2-x}{2}\right) \sin \left(\frac{2+x-(2-x)}{2}\right)}{x} = A \cos B$.$\lim _{x \rightarrow 0} \frac{2 \cos (2) \sin (x)}{x} = A \cos B$.Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we get:$2 \cos 2 = A \cos B$.Comparing both sides,we find $A = 2$ and $B = 2$.

If $\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=A \cos B$,then the values of $A$ and $B$ respectively are

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