A) \[\sin a\]
B) \[\cos a\]
C) \[-\sin a\]
D) \[\frac{1}{2}\cos a\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,2\,\sin \,a\,.\,\frac{(\cos x-1)}{x\sin x}\]\[=-2\,\sin a\,.\,\frac{(1-\cos x)}{{{x}^{2}}}\,.\,\left( \frac{x}{\sin x} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,-2\sin a\,.\,\frac{2\,{{\sin }^{2}}(x/2)}{4\,{{\left( \frac{x}{2} \right)}^{2}}\,\left( \frac{\sin x}{x} \right)}=-\sin a\].
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