How to find limit $\lim \limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $ I need to find this limit $\lim\limits_{ x\rightarrow +\infty  }{ \tan { { \left( \frac { \pi x }{ 2x+1 }  \right)  }^{ 1/x } }  } $.
Give a hint please.Thanks
 A: Hints:
$$\frac{\pi x}{2x+1} = \frac{\pi}{2}-\frac{\pi}{4x+2}\tag{1}$$
$$\tan\left(\frac{\pi x}{2x+1}\right) = \cot\left(\frac{\pi}{4x+2}\right)\tag{2}$$
$$\frac{1}{x}\,\log\cot\left(\frac{\pi}{4x+2}\right)=\frac{\log\frac{4}{\pi}+\log x}{x}+O\left(\frac{1}{x^2}\right)\text{ as }x\to+\infty.\tag{3}$$
A: We can do that using equivalents.
It amounts to finding the limit of
$\;\displaystyle \frac1x\ln\biggl(\tan\Bigl(\frac{\pi x}{2x+1}\Bigr)\biggl)$.
Set $u=\dfrac1x$. We have
$$\frac{\pi x}{2x+1}=\frac\pi2\frac1{1+\dfrac2x}=\frac\pi2\frac1{1+2u}=\frac\pi2\bigl(1-2u+o(u)\bigr)=\frac\pi2-\pi u+o(u) $$
Henceforth
$$\tan\Bigl(\frac{\pi x}{2x+1}\Bigr)\biggl)=\tan\Bigl(\frac\pi2-\pi u+o(u)\Bigr)=\frac1{\tan\bigl(\pi u-o(u)\bigr)}\sim_0\frac1{\pi u}.$$
As $\dfrac1{\pi u}$ does not approach $1$, we may take the logarithms:
$\;\ln\biggl(\tan\Bigl(\frac{\pi x}{2x+1}\Bigr)\biggl)\sim -\ln(\pi u)$, 
and finally $$\frac1x\ln\biggl(\tan\Bigl(\dfrac{\pi x}{2x+1}\Bigr).\biggl)\sim u\ln(\pi u)\xrightarrow[u\to 0]{} 0$$
A: Let $y$ be the limit . Take logarithm both the sides 
$$\log y =\lim_{x \to \infty} \frac{1}{x}\log(tan(\frac{\pi x}{2x+1}))$$
This is infinity by infinity form , apply L-Hospital rule.
$$\log y= \lim_{x \to \infty} \frac{\pi \cot(\frac{\pi x }{2x+1})}{(2x+1)^2}$$
Put limit 
$$\log y =0$$
Thus,
$$y=1$$
