A) \[{{e}^{\frac{1}{\pi }}}\]
B) \[{{e}^{\frac{2}{\pi }}}\]
C) \[{{e}^{\pi }}\]
D) \[{{e}^{\frac{3}{\pi }}}\]
Correct Answer: B
Solution :
\[\underset{x\,\to \,a}{\mathop{\lim }}\,{{\left( 2-x \right)}^{\tan \frac{\pi x}{2}}}~~~~~~\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{1}^{\infty }}\,\,form \right]\] \[\left[ \underset{x\,\to \,a}{\mathop{lim}}\,\,\,f(x)=1 and \underset{x\,\to \,a}{\mathop{lim}}\,\,\,g(x)=\,\,\infty \right.\] \[\left. \Rightarrow \,\,\underset{x\to a}{\mathop{\lim }}\,\,\,f{{(x)}^{g(x)}}={{e}^{\underset{x\to a}{\mathop{\lim }}\,g(x)(f(x)-1)}} \right]\] \[\left. \Rightarrow \,\,\underset{x\to 1}{\mathop{\lim }}\,\,\,{{(2-x)}^{tan\frac{\pi x}{2}}}\,\,=\,\,{{e}^{\underset{x\to a}{\mathop{\lim }}\,\,\,\tan \frac{\pi x}{2}(2-x-1)}} \right]\] \[\Rightarrow \,\,\,\,\,{{e}^{\underset{x\to 1}{\mathop{\lim }}\,}}^{tan\frac{\pi x}{2}(1-x)}\,\,=\,\,{{e}^{\underset{x\to 1}{\mathop{\lim }}\,\frac{1-x}{\cot \left( \frac{\pi }{2}x \right)}}}\] [Using L-Hospital rule] \[=\,{{e}^{\frac{2}{\pi }}}\]You need to login to perform this action.
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