MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (2 + x) - \sin (2 - x)}}{x} = $
- A$\sin 2$
- B$2\sin 2$
- ✓$2\cos 2$
- D$2$
$i.e.,$ $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (2 + x) - \sin (2 - x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2\cos 2.\sin x}}{x}$
$ = 2\cos 2.\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 2\cos 2$
You may also apply $L-$ Hospital rule.
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