\[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\sin x-\sin \alpha }{x-\alpha }=\]

\[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\sin x-\sin \alpha }{

1.	\[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\sin x-\sin \alpha }{x-\alpha }=\]
A.	0
B.	1
C.	\[\sin \alpha \]
D.	\[\cos \alpha \]
Answer» E.

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\[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{\sin x-\sin \alpha }{x-\alpha }=\]

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