A) zero
B) \[\infty \]
C) 1
D) cannot be determined
Correct Answer: A
Solution :
Given, \[\underset{x\to 0}{\mathop{\lim }}\,x\log \sin x\] \[=\underset{x\to 0}{\mathop{\lim }}\,x\frac{\log \sin x}{1/x}\] \[\left( \frac{\infty }{\infty }form \right)\] By using L Hospitals rule, we get \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(1/\sin x)\cos x}{-1/{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,-\frac{\cos x.x}{\frac{\sin x}{x}}\] \[=\frac{0}{1}=0\] \[\therefore \]Required limit is 0.
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