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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