A) 0
B) \[\infty \]
C) ?1
D) 1
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\tan }^{-1}}x}{x}\], \[\left( \frac{0}{0}\,\text{form} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\frac{1}{1+{{x}^{2}}}}{1}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{1+{{x}^{2}}}=\frac{1}{1+0}=1\].
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