A) 2
B) \[\frac{1}{2}\]
C) 0
D) None of these
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\tan }^{-1}}2x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{2.\frac{1}{1+4{{x}^{2}}}}\] (by L-Hospital rule) \[=\frac{1}{2}\]You need to login to perform this action.
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