I have to solve the limit with L'hopital's rule, but I can't. I understand that since there is a power here, I need to take a log from it.
Limit: $\lim _{x \rightarrow \infty} \sqrt[x]{\tan \left(\frac{\pi x}{2 x+1}\right)}$
After i taking a log, i get $e^{\lim _{x \rightarrow \infty} \frac{\ln \left(\tan \left(\frac{\pi x}{2 x+1}\right)\right)}{x}}$ which is $(\frac{\infty}{\infty})$, so i can use L'hospital's rule.
After I take derivatives and simplify them, I get such a limit:
$e^{\lim _{x \rightarrow \infty} \frac{\pi}{\cos \left(\frac{\pi x}{2 x+1}\right)(2 x+1)^{2}}}$
However, here I can no longer apply the L'hopital's rule anymore, but the indeterminate form hasn't disappeared since cos -> cos($\pi/2$) = 0, and $(2x+1)^2$ -> +inf.
I don't know how to proceed. Maybe I chose the wrong approach from the very beginning?