Limit of $\left(\frac{-2}{\pi}\cdot\arctan{x}\right)^x$ when $x\to -\infty$, without using l'Hôpital
Since $x\lt0$, $\arctan x=-\arctan(1/x)-\pi/2$ hence you are looking at $(1+u(x))^x$ with $u(x)=(2/\pi)\arctan(1/x)\sim2/(\pi x)$. For every $a$, $(1+a/x)^x\to\mathrm e^a$ when $x\to\infty$ hence the quantity you are looking at converges to $\mathrm e^{2/\pi}$.