Limit of $\left(\frac{-2}{\pi}\cdot\arctan{x}\right)^x$ when $x\to -\infty$, without using l'Hôpital Can somebody help me find this limit without using the l'Hôpital's rule. I'm probably overseeing something really simple but all I can get are $0/0$ forms or such:
$$\lim_{x\to -\infty}\left(\frac{-2}{\pi}\cdot\arctan{x}\right)^x$$
 A: 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 $\mathrm e^{2/\pi}$.
A: Let $\frac{-2}{\pi}\cdot\arctan x=1-y$
$\implies \arctan x=-(\frac\pi2-\frac\pi 2y)$
$\implies x=\tan \{-(\frac\pi2-\frac\pi 2y)\}=-\tan(\frac\pi2-\frac\pi 2y)=-\cot \frac\pi2y$ 
and as  $x\to-\infty\implies\cot \frac\pi2y=\infty\implies y\to 0$
$$\lim_{x\to -\infty}\left(\frac{-2}{\pi}\cdot\arctan{x}\right)^x$$
$$=\lim_{y\to0}(1-y)^{-\cot \frac{\pi y}2}$$
$$=(\lim_{y\to0}(1-y)^{-y})^{\frac{\lim_{y\to0}\cos\frac{\pi y}2}{\lim_{y\to0}\frac{\sin \frac{\pi y}2}y}}=e^{\frac2\pi}$$
A: Let $\arctan x=y-\frac\pi2\implies x=\tan(y-\frac\pi2)=-\tan(\frac\pi2-y)=-\cot y$
and $x\to-\infty, \cot y=\infty\implies y\to 0$
$$\lim_{x\to -\infty}\left(\frac{-2}{\pi}\cdot\arctan{x}\right)^x$$
$$=\lim_{y\to 0}\{\frac{-2}{\pi}(y-\frac\pi2)\}^{-\cot y}=\lim_{y\to 0}\left(1-\frac y{\frac\pi2}\right)^{-\cot y}$$
$$z=\left(1-\frac y{\frac\pi2}\right)^{-\cot y}$$
$$\implies \log z=-\cot y\log\left(1-\frac y{\frac\pi2}\right)=\frac{\cos y}{\left(\frac{-\sin y}{-y}\right)}\frac{\log\left(1-\frac y{\frac\pi2}\right)}{\frac{-y}{\frac\pi2}}\frac2\pi$$
So, $$\lim_{y\to 0}\log z=\frac2\pi \frac{\lim_{y\to 0}\cos y}{\lim_{y\to 0}\left(\frac{-\sin y}{-y}\right)}\cdot \lim_{y\to 0}\frac{\log\left(1-\frac y{\frac\pi2}\right)}{\frac{-y}{\frac\pi2}}=\frac2\pi$$
So,  $$\lim_{y\to 0}z=e^{\frac2\pi}$$
