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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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The correct answer is:
2,2
We have,
$\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=A \cos B$
$\lim _{x \rightarrow 0} \frac{2 \cos \left(\frac{2+x+2-x}{2}\right) \sin \left(\frac{2+x-2+x}{2}\right)}{x}$
$\begin{aligned} & =\lim _{x \rightarrow 0} 2 \cos 2 \frac{\sin x}{x} \\ & =2 \cos 2 \lim _{x \rightarrow 0} \frac{\sin x}{x}=2 \cos 2\end{aligned}$
Thus, $A=2$ and $B=2$
$\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=A \cos B$
$\lim _{x \rightarrow 0} \frac{2 \cos \left(\frac{2+x+2-x}{2}\right) \sin \left(\frac{2+x-2+x}{2}\right)}{x}$
$\begin{aligned} & =\lim _{x \rightarrow 0} 2 \cos 2 \frac{\sin x}{x} \\ & =2 \cos 2 \lim _{x \rightarrow 0} \frac{\sin x}{x}=2 \cos 2\end{aligned}$
Thus, $A=2$ and $B=2$
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