A) \[2\]
B) \[\frac{1}{2}\]
C) \[-2\]
D) \[\frac{-1}{2}\]
Correct Answer: A
Solution :
We have,\[\underset{x\to 1}{\mathop{\lim }}\,\frac{\tan ({{x}^{2}}-1)}{x-1}\left( \frac{0}{0}form \right)\]form \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{\sec }^{2}}({{x}^{2}}-1).2x}{1}\] [using L? Hospital?s rule] \[=2.{{\sec }^{2}}(0)=2\]
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