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Calculate  \mathrm{\lim _{x \rightarrow \infty}\left(\frac{2}{\pi} \arctan (x)\right)^{\frac{x^2}{1+2 x}}}

Option: 1

1


Option: 2

-2


Option: 3

0


Option: 4

5


Answers (1)

best_answer

We are to compute 

              \mathrm{\lim _{x \rightarrow \infty} \frac{x^2}{1+2 x} \log \left(\frac{2}{\pi} \tan ^{-1} x\right)}

With the change of variable \mathrm{=1-\frac{2}{\pi} \tan ^{-1} x, \text { then }}

\mathrm{\lim _{x \rightarrow \infty} \frac{x^2}{1+2 x} \log \left(\frac{2}{\pi} \tan ^{-1} x\right)=\lim _{u \rightarrow 0^{+}} \frac{\cot ^2 \frac{\pi}{2} u}{1+2 \cot \frac{\pi}{2} u} \log (1-u)}

                                                              \mathrm{=\lim _{u \rightarrow 0^{+}} \frac{2}{\pi} \cdot \frac{\frac{\pi}{2} u}{\sin \frac{\pi}{2} u} \cdot \frac{\cos ^2 \frac{\pi}{2} u}{\sin \frac{\pi}{2} u+2 \cos \frac{\pi}{2} u} \cdot \frac{1}{u} \cdot \log (1-u)}

Note that

                \mathrm{\lim _{u \rightarrow 0^{+}} \frac{1}{u} \cdot \log (1-u)=-\lim _{u \rightarrow 0^{+}} \frac{1}{u} \int_0^u \frac{1}{1-t} d t=-\lim _{u \rightarrow 0^{+}} \frac{1}{1-\eta_u}=-1,}

where \mathrm{\eta_u} is in between u and 0 , chosen by Mean Value Theorem.

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chirag

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