#### Calculate  $\mathrm{\lim _{x \rightarrow \infty}\left(\frac{2}{\pi} \arctan (x)\right)^{\frac{x^2}{1+2 x}}}$Option: 1 1Option: 2 -2Option: 3 0Option: 4 5

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.