How to find limit $\lim \limits_{ x\rightarrow +\infty }{ \tan { { \left( \frac { \pi x }{ 2x+1 } \right) }^{ 1/x } } } $

Question

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

Answer 1

Hints: $$\frac{\pi x}{2x+1} = \frac{\pi}{2}-\frac{\pi}{4x+2}\tag{1}$$ $$\tan\left(\frac{\pi x}{2x+1}\right) = \cot\left(\frac{\pi}{4x+2}\right)\tag{2}$$ $$\frac{1}{x}\,\log\cot\left(\frac{\pi}{4x+2}\right)=\frac{\log\frac{4}{\pi}+\log x}{x}+O\left(\frac{1}{x^2}\right)\text{ as }x\to+\infty.\tag{3}$$

Answer 2

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 $$\frac1x\ln\biggl(\tan\Bigl(\dfrac{\pi x}{2x+1}\Bigr).\biggl)\sim u\ln(\pi u)\xrightarrow[u\to 0]{} 0$$

Answer 3

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, $$y=1$$