Trigonometric limit with L'hopital's rule and indeterminate form $\infty^{0}$ 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?
 A: I developed it as follows. Please, doublecheck if what I have done is fine.
\begin{eqnarray}
\mathcal L &=& \lim_{x\to +\infty} \sqrt[x]{\tan\left(\frac{\pi x}{2x+1}\right)}=\\
&=&\lim_{x\to +\infty}\left[\tan \left(\frac{\pi}2-\frac{\pi}{4x+2}\right)\right]^{\frac1x}=\\
&=&\lim_{x\to+\infty}\left[\frac{\sin\left(\frac{\pi}2-\frac{\pi}{4x+2}\right)}{\cos\left(\frac{\pi}2-\frac{\pi}{4x+2}\right)}\right]^{\frac1x}=\\
&=&\lim_{x\to+\infty}\left[\frac{\cos\left(\frac{\pi}{4x+2}\right)}{\sin\left(\frac{\pi}{4x+2}\right)}\right]^{\frac1x}=\\
&=&e^{\lim_{x\to+\infty}\frac1x\log\left[\frac{\cos\left(\frac{\pi}{4x+2}\right)}{\sin\left(\frac{\pi}{4x+2}\right)}\right]}=\\
&=&e^{\lim_{x\to+\infty}\frac1x\log\left(\frac{4x+2}{\pi}\right)}=1,
\end{eqnarray}
where I used
$$\cos\left(\frac{\pi}{4x+2}\right) \to 1$$
and
$$\sin\left(\frac{\pi}{4x+2}\right)\sim \frac{\pi}{4x+2},$$
for $x\to+\infty$.
A: Avoid writing complicated exponents and first do the substitution $x=1/t$, so the limit of the logarithm becomes
$$
\lim_{t\to0^+}t\log\tan\dfrac{\pi t}{2+t}
$$
For small enough $t$, this function is negative. Also, for small enough $\alpha$,
$$
\tan\alpha>\alpha
$$
so we have
$$
t\log\frac{\pi t}{2+t}\le t\log\tan\dfrac{\pi t}{2+t}\le 0
$$
Now
$$
t\log\frac{\pi t}{2+t}=t\log\pi+t\log t-t\log(2+t)
$$
and
$$
\lim_{t\to0^+}t\log\frac{\pi t}{2+t}=
\lim_{t\to0^+}(t\log\pi+t\log t-t\log(2+t))=0
$$
By squeezing, also
$$
\lim_{t\to0^+}t\log\tan\dfrac{\pi t}{2+t}=0
$$
and therefore your limit is $e^0=1$.
A: Or you could just use $\tan\frac{\pi x}{2x+1}=\cot\frac{\pi}{4x+2}\sim\frac{4x}{\pi}$ to prove the limit is $1$, viz.$$\lim_{x\to\infty}\frac{\ln(4x/\pi)}{x}=\lim_{x\to\infty}\frac{1}{x}=0.$$
A: Write ´
$$e^{\lim_{x\to \infty}\frac{\ln\left(\tan\left(\frac{\pi x }{2x+1}\right)\right)}{x}}$$
A: $$\lim_{x\to\infty}\left(\tan\dfrac{\pi x}{2x+1}\right)^{1/x}$$
$$=\left(\lim_{x\to\infty}\left(\tan\dfrac{\pi x}{2x+1}\right)^{\dfrac1{\tan\dfrac{\pi x}{2x+1}}}\right)^{\lim_{x\to\infty}\dfrac{\tan\dfrac{\pi x}{2x+1}}x}$$
For the inner limit, set $\tan\dfrac{\pi x}{2x+1}=n$
and use  How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?
For the exponent, set $1/x=h$ 
$$\lim_{x\to\infty}\dfrac{\tan\dfrac{\pi x}{2x+1}}x=\lim_{h\to0^+}h\tan\dfrac{\pi}{2+h}=\lim_{h\to0^+}h\cot\dfrac{\pi h}{h+2}=\lim_{h\to0^+}\cos\dfrac{\pi h}{h+2}\cdot\lim_{h\to0^+}\dfrac h{\sin\dfrac{\pi h}{h+2}}=?$$
