A) 0
B) 1
C) \[\sin \alpha \]
D) \[\cos \alpha \]
Correct Answer: D
Solution :
\[\underset{x\to \alpha }{\mathop{\lim }}\,\,\,\frac{\sin x-\sin \alpha }{x-\alpha }\] \[\underset{x\to \alpha }{\mathop{\lim }}\,\,\,\frac{\cos x}{1}=\cos \alpha \], (Apply L-Hospital's rule)
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