$ \lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x$ and $\lim_{x\rightarrow 3^+} \frac{\cos x \ln(x-3)}{\ln(e^x-e^3)}$? 
I got stuck on two exercises below
  $$
\lim\limits_{x\rightarrow +\infty} \left(\frac{2}{\pi} \arctan x \right)^x   \\
\lim_{x\rightarrow 3^+} \frac{\cos x  \ln(x-3)}{\ln(e^x-e^3)}
$$ 

For the first one , let $y=(\frac{2}{\pi} \arctan x )^x $, so $\ln y =x\ln (\frac{2}{\pi} \arctan x )$, the right part is $\infty \cdot 0$ type, but seemly, the L 'hopital's rule is useless.  PS: I know the $\infty \cdot 0$ can be become to $\frac{\infty}{\infty}$ or $\frac{0}{0}$. But when I use the L 'hopital's rule to the $\frac{\infty}{\infty}$ or $\frac{0}{0}$ the calculation is complex and useless.
For the second one , it is $\frac{\infty}{\infty}$ type, also useless the L 'hopital's rule is. How to calculate it ? 
 A: Rewrite $\infty\cdot 0$ as $\infty \cdot \dfrac{1}{\infty}$. Now you can apply L'Hopital's rule: $$\lim_{x\to +\infty}\dfrac{\left(\ln 2/\pi\cdot\arctan x \right)}{1/x}=\lim_{x\to +\infty}\dfrac{\pi/2\cdot \arctan x}{-1/x^2}\cdot \dfrac{1}{1+x^2}=-\dfrac{\pi }{2}\lim_{x\to +\infty}\arctan x\cdot \dfrac{x^2}{1+x^2}$$
A: Without L'Hospital
$$y=\left(\frac{2}{\pi} \arctan (x) \right)^x\implies \log(y)=x \log\left(\frac{2}{\pi} \arctan (x) \right)  $$
Now, by Taylor for large values of $x$
$$\arctan (x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$\frac{2}{\pi} \arctan (x) =1-\frac{2}{\pi  x}+\frac{2}{3 \pi  x^3}+O\left(\frac{1}{x^4}\right)$$ Taylor again
$$\log\left(\frac{2}{\pi} \arctan (x) \right)= -\frac{2}{\pi  x}-\frac{2}{\pi ^2 x^2}+O\left(\frac{1}{x^3}\right)$$
$$\log(y)=x\log\left(\frac{2}{\pi} \arctan (x) \right)= -\frac{2}{\pi  }-\frac{2}{\pi ^2 x}+O\left(\frac{1}{x^2}\right)$$ Just continue with Taylor using $y=e^{\log(y)}$ if you want to see not only the limit but also how it is approached
A: You can solve the first one using


*

*$\arctan x + \operatorname{arccot}x = \frac{\pi}{2}$

*$\lim_{y\to 0}(1-y)^{1/y} = e^{-1}$

*$x\operatorname{arccot}x \stackrel{\stackrel{x =\cot u}{u\to 0^+}}{=} \cot u\cdot u = \cos u\cdot \frac{u}{\sin u} \stackrel{u \to 0^+}{\longrightarrow} 1$
\begin{eqnarray*} \left(\frac{2}{\pi} \arctan x \right)^x
& \stackrel{\arctan x = \frac{\pi}{2}-\operatorname{arccot}x}{=} & \left( \underbrace{\left(1- \frac{2}{\pi}\operatorname{arccot}x\right)^{\frac{\pi}{2\operatorname{arccot}x}}}_{\stackrel{x \to +\infty}{\longrightarrow} e^{-1}} \right)^{\frac{2}{\pi}\underbrace{x\operatorname{arccot}x}_{\stackrel{x \to +\infty}{\longrightarrow} 1}} \\
& \stackrel{x \to +\infty}{\longrightarrow} & e^{-\frac{2}{\pi}}
\end{eqnarray*}
The second limit is quite straight forward as $\lim_{x\to 3+}\cos x = \cos 3$. Just consider 


*

*$\frac{\ln(x-3)}{\ln(e^x-e^3)}$ and apply L'Hospital.

A: For the first: taking $\log$ and doing the cov $x = 1/t$ and using L'Hôpital:
$$
\lim_{x\to+\infty}x\log\left(\frac{2}{\pi}\arctan x \right) =
\lim_{t\to 0^+}\frac 1t\log\left(\frac{2}{\pi}\arctan(1/t)\right) =
\lim_{t\to 0^+}\frac {-1}{(t^2 + 1)\arctan(1/t))} = -\frac 2\pi.
$$
