How to solve $ \lim_{x \to \infty} \Big(\frac{\pi}{2}\arctan x\Big)^x$ I am not sure what should I use. series for $\arctan x$ or L'hopital's rules cause it converges to $\infty$
$$\lim_{x \to \infty} \Big(\frac{\pi}{2}\arctan x\Big)^x$$
 A: $$ \lim_{x \to \infty} \left(\frac2\pi\arctan x\right)^x=\lim_{t\to 0^+} \left(\frac2\pi\text{arccot} t\right)^{1/t}=\lim_{t\to 0^+} \left(1-\frac2\pi\arctan t\right)^{1/t}\\
=\lim_{u\to 0^+} \left(1-\frac2\pi u\right)^{1/\tan u}
=\lim_{u\to 0^+} \left(\left(1-\frac2\pi u\right)^{1/u}\right)^{u/\tan u}\\
=e^{-2/\pi}.$$
A: When $x\to +\infty$ we have that $\arctan x \to \frac {\pi} 2$ thus $\lim_{x\to +\infty}\left(\frac {\pi^2} 4\right)^x=+\infty$ because $\frac {\pi^2} 4 \gt \frac {3^2} 4 \gt1$
A: Assuming we are looking at $[(2/\pi)\arctan x]^x,$ apply $\ln$ to get
$$x\ln [(2/\pi)\arctan x] = \frac{\ln [(2/\pi)\arctan x]}{1/x}.$$
As $x\to \infty,$ the numerator and denominator both $\to 0,$ so let's try L'Hopital. The quotient of derivatives equals
$$\tag 1\frac{1}{(2/\pi)\arctan x}\cdot\frac{2/\pi}{1+x^2}\cdot\frac{1}{-1/x^2}.$$
The first fraction $\to 1,$ while the product of the last two fractions $\to -2/\pi.$ Thus the limit of $(1)$ is $-2/\pi.$ Exponentiating back then gives $e^{-2/\pi}$ for the original limit.
