<s>[INST] $ \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 ? 
 [/INST] 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
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