find this exponential limit I tried turning it into an exp limit and using l'Hopital but I must have done some mistake because the limit to me results infinite whereas the answer is -1, can someone give a hint please
 

 A: Oh cool. This is very tricky I think.
$\lim_{x \to \infty} (\pi/2 - \arctan(x))^{1/\ln(x)}$
$= (LHR) \exp(\lim_{x \to \infty} \frac{\ln(\pi/2 - \arctan(x))}{\ln(x)})$
$= \exp(\lim_{x \to \infty} \frac{1/(\pi/2 - \arctan(x)) (\frac{-1}{x^2+1})}{1/(x)})$
$= \exp(\lim_{x \to \infty} \frac{(\frac{-x}{x^2+1})}{\pi/2 - \arctan(x)})$
$= \exp(\lim_{x \to \infty} \frac{(\frac{-x}{x^2+1})}{\pi/2 - \arctan(x)})$
$= \exp(\lim_{x \to \infty} \frac{(\frac{-x}{x^2+1})}{\pi/2 - \arctan(x)})$
$= (LHR) \exp(-\lim_{x \to \infty} \frac{(x^2 - 1)/(x^2 + 1)^2}{1/(x^2 + 1)})$ (The details I leave to you)
$= \exp(-\lim_{x \to \infty} \frac{(x^2 - 1)}{(x^2 + 1)})$
$= 1/e$
A: Why don't you start using the identity $$\tan^{-1}(x)+\tan^{-1}(\frac 1x)=\frac {\pi}2$$ $$A=\Big(\frac{\pi}2 - \tan^{-1}(x)\Big)^{\frac 1{\ln(x)}}=\Big(\tan^{-1}(\frac 1x)\Big)^{\frac 1{\ln(x)}}$$ Now, taking logarithms $$\log(A)=\frac{\log(\tan^{-1}(\frac 1x))}{\log(x)}$$ and set $u=\log(\tan^{-1}(\frac 1x))$, $v=\log(x)$. Now $$u'=-\frac{1}{\left(1+x^2\right)  \tan ^{-1}\left(\frac{1}{x}\right)}$$ with $v'=\frac{1}{x}$. So, $$A=\frac{u'}{v'}=-\frac{x}{\left(1+x^2\right)  \tan ^{-1}\left(\frac{1}{x}\right)}$$ and you probably know that, for small $y$, $\tan^{-1}(y)\approx y$; so, since $x$ is large, replace $y$ by $\frac{1}{x}$ to get $$\log(A)=-\frac{x^2}{1+x^2}$$ So, for an infinite value of $x$, the limit of $\log(A)$ is $-1$ and then the limit of $A$ is $\frac 1e$.
