A) \[{{e}^{\frac{1}{\pi }}}\]
B) \[{{e}^{-\frac{1}{\pi }}}\]
C) \[{{e}^{\frac{2}{\pi }}}\]
D) \[{{e}^{-\frac{2}{\pi }}}\]
Correct Answer: A
Solution :
[a] \[L=\underset{x\to 1}{\mathop{\lim }}\,{{\left( \frac{4}{\pi }{{\tan }^{-1}}x \right)}^{\frac{1}{({{x}^{2}}-\,1)}}}={{e}^{\underset{x\to 1}{\mathop{\lim }}\,\,\frac{1}{{{x}^{2}}-\,1}\left( \frac{4}{\pi }{{\tan }^{-1}}x-\,t \right)}}\] Now, \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( \frac{4}{\pi }{{\tan }^{-1}}x-1 \right)}{{{x}^{2}}-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{4}{\pi }\frac{1}{1+{{x}^{2}}}}{2x}=\frac{1}{\pi }\] \[\therefore \,\,\,L={{e}^{\frac{1}{\pi }}}\]
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