Limit problem arctg (1 ^ infinty) Can anyone help me with this limit problem without L'Hopital rule and Taylor series?
$$\lim_{x\rightarrow\ -\infty}\left(-2\frac{\arctan(x)}{\pi}\right)^x$$
 A: Let $y=\arctan x$, so your limit is as $y\to-\pi/2$, and the exponent is $\tan y$.
Let $z=y+\pi/2$, so your limit is as $z\to0^+$.
Use the fact that $(1-w)^{-w}\to e$ as $w\to 0$
A: For $x<0$, we have
$$
\arctan x+\arctan\frac{1}{x}=-\frac{\pi}{2}
$$
so
$$
-\frac{2}{\pi}\arctan x=1+\frac{2}{\pi}\arctan\frac{1}{x}
$$
Now, set $t=1/x$ and your limit becomes
$$
\lim_{t\to0^-}\left(1+\frac{2}{\pi}\arctan t\right)^{1/t}
$$
and here you can first compute the limit of the logarithm:
$$
\lim_{t\to0^-}\frac{\log(1+\frac{2}{\pi}\arctan t)}{t}
$$
which is the derivative at $0$ of $f(t)=\log(1+\frac{2}{\pi}\arctan t)$; since
$$
f'(t)=\frac{1}{1+\frac{2}{\pi}\arctan t}\frac{2}{\pi}\frac{1}{1+t^2}
$$
we have
$$
f'(0)=\frac{2}{\pi}
$$
So your limit is
$$
e^{2/\pi}
$$
A: We can find the limit of $\lim_{x\to -\infty} \ln \left(\frac{-2\arctan x }{\pi}\right)^x=\lim_{x\to -\infty} x\ln \left(\frac{-2\arctan x }{\pi}\right)=\lim_{x \to -\infty} \frac {\ln \left(\frac{-2\arctan x }{\pi}\right)}{1/x}$.
It is well known that $\lim_ \limits{x \to -\infty} \arctan x=\frac{-\pi}{2}.$ 
Using this fact, we now have an indeterminate limit, so we apply L'Hopital's rule. $\lim_{x \to -\infty} \frac {\ln \left(\frac{-2\arctan x }{\pi}\right)}{1/x}= \lim_{x \to -\infty}\frac{\frac{1}{ \arctan x (1+x^2)}}{-1/x^2}.$
Rearranging the fraction gives us $\lim_{x \to -\infty}{\frac{-x^2}{ \arctan (x) (1+x^2)}}.$ Dividing both the numerator and denominator by $x^2$ gives:
$$\lim_{x \to -\infty} \frac{-1}{\arctan(x)(1/x^2+1)}.$$ Now, the limit can be safely be evaluated as $\frac{2}{\pi}.$ 
However, remember that we started by taking the natural logarithm of the initial limit. Thus, we must exponentiate with a base of $e$. Thus:

$$\lim_ \limits{x\to -\infty} \left(\frac{-2\arctan x }{\pi}\right)^x=e^{2/\pi}$$

A: I thought it might be instructive to see an approach that does not rely on differential calculus, but rather standard inequalities and the squeeze theorem.  
I showed in THIS ANSWER and THIS ONE , using only (1) the limit definition for the exponential function and (2) Bernouli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{1+x}\le \log(1+x)\le x} \tag 1$$ 
for $x>-1$.  

Next, recall from basic geometry that the sine function and cosine functions satisfy the inequalities
$$x\le \sin (x)\le x\cos(x) \tag 2$$
for $-\pi/2 \le x\le 0$.  It is straightforward to show from $(2)$ that the arctangent function satisfies the inequlities
$$\bbox[5px,border:2px solid #C0A000]{\frac1x\le \arctan(1/x)\le -\frac{1}{\sqrt{1+x^2}}} \tag 3$$
for $x<0$.
We will use the inequalities in $(1)$ and $(3)$ in the ensuing analysis.

To begin, we write for $x<0$
$$\begin{align}
\left(-\frac2\pi \arctan(x)\right)^x&=\exp\left(x\log\left(-\frac2\pi \arctan(x)\right)\right)\\\\
&=\exp\left(x\log\left(1+\frac2\pi \arctan(1/x)\right)\right)\tag 4
\end{align}$$
Let's bound the argument of the exponential function on the right-hand side of $(4)$ using $(1)$ and $(3)$.  The upper bound provided by $(1)$ and $(3)$ reveals
$$\begin{align}
x\log\left(1+\frac2\pi \arctan(1/x)\right)&\le x\left(\frac2\pi \arctan(1/x)\right)\\\\
&\le- \frac2\pi \frac{x}{\sqrt{x^2+1}} \tag 5
\end{align}$$
while the lower bound provide by $(1)$ and $(3)$ reveals
$$\begin{align}
x\log\left(1+\frac2\pi \arctan(1/x)\right)&\ge x\left(\frac2\pi \frac{\arctan(1/x)}{1+\arctan(1/x)}\right)\\\\
&
\ge \frac2\pi \frac{\sqrt{1+x^2}}{1+\sqrt{1+x^2}} \tag 6
\end{align}$$ 
Putting $(5)$ and $(6)$ together yields
$$\frac2\pi \frac{\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\le x\log\left(1+\frac2\pi \arctan(1/x)\right)\le -\frac2\pi \frac{x}{\sqrt{x^2+1}}$$
whereupon applying the squeeze theorem, we obtain the limit
$$\lim_{x\to -\infty}x\log\left(1+\frac2\pi \arctan(1/x)\right)=\frac2\pi$$
Finally, using continuity of the exponential function, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to -\infty}\left(-\frac2\pi \arctan(x)\right)^x=e^{2/\pi}}$$
