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MCQ

$\mathop {\lim }\limits_{x \to \alpha } \frac{{\sin x - \sin \alpha }}{{x - \alpha }} = $

A
$0$
B
$1$
C
$\sin \alpha $
✓
$\cos \alpha $

✓

Answer

Correct option: D.

$\cos \alpha $

d
(d) $\mathop {\lim }\limits_{x \to \alpha } \,\,\frac{{\sin x - \sin \alpha }}{{x - \alpha }}$

$\mathop {\lim }\limits_{x \to \alpha } \,\,\frac{{\cos x}}{1} = \cos \alpha $, (Apply $L-$Hospital's rule)

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