calculating a limit of sequence with tan Calculating a limit
$$\lim_{n\longrightarrow +\infty} \dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}$$
thanks.
 A: *

*Change the variable $n \to \frac 1n$
$$
L = \lim_{n \to 0^+} \frac {\tan \left ( \frac {\frac \pi n}{\frac 2n+1}\right )}{\sqrt[3]{\frac 1{n^3}+\frac 2n - 1}} = \lim_{n \to 0^+} \frac {n \tan \left ( \frac \pi{2+n} \right )}{\sqrt[3]{1+2n^2-n^3}} = \lim_{n \to 0^+} \left [ n \tan \left( \frac \pi{2+n} \right ) \right ]
$$

*Use $\tan \alpha = \frac 1{\tan \left(\frac \pi 2 - \alpha \right )}$
$$
L = \lim_{n \to 0^+} \frac n{\tan \left ( \frac \pi 2 - \frac \pi{2+n} \right )}  = \lim_{n \to 0^+} \frac n{\tan \left( \frac {\pi n}{4+2n}\right )} = \lim_{n \to 0^+} \frac n{\frac {\pi n}{4+2n}} = \lim_{n \to 0^+} \frac {4+2n}\pi = \frac 4\pi
$$

*Check - WA.

A: Hint : $cos(\frac{\pi}{2}+x)$ is equivalent to $-x$ when $ x \to 0$ and $\frac{\pi n}{2n+1}=\frac{\pi}{2}(1-\frac{1}{2n+1} )$
A: First write
$$\tan\left(\frac{\pi n}{2n+1}\right)=\cot\left(\frac{\pi/2}{2n+1}\right)$$
Then, in THIS ANSWER I showed that from basic geometry the cotangent function satisfies the inequalities
$$\frac{\cos(x)}{x}\le \cot(x)\le \frac1x$$
for $0 < x\le \pi/2$.  
Therefore, we have
$$\frac{(2n+1)\cos\left(\frac{\pi/2}{2n+1}\right)}{\frac\pi 2 n\left(1+\frac2{n^2}+\frac{1}{n^3}\right)^{1/3}}\le \frac{\tan\left(\frac{\pi n}{2n+1}\right)}{\left(n^3+2n-1\right)^{1/3}}\le\frac{2n+1}{\frac\pi 2 n\left(1+\frac2{n^2}+\frac{1}{n^3}\right)^{1/3}}$$
whereupon invoking the Squeeze Theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\tan\left(\frac{\pi n}{2n+1}\right)}{\left(n^3+2n-1\right)^{1/3}}=\frac{4}{\pi}}$$
A: Consider that
$$
\lim_{n\to\infty}\frac{n}{\sqrt[3]{n^3+2n-1}}=
\lim_{n\to\infty}\frac{n}{n\sqrt[3]{1+\dfrac{2}{n^2}-\dfrac{1}{n^3}}}=1
$$
so what you really need to compute is
$$
\lim_{n\to\infty}\frac{\tan\dfrac{\pi n}{2n+1}}{n}=
\lim_{n\to\infty}\frac{\sin\dfrac{\pi n}{2n+1}}{n}
  \frac{1}{\cos\dfrac{\pi n}{2n+1}}
$$
Since
$$
\sin\dfrac{\pi n}{2n+1}\xrightarrow{n\to\infty}1
$$
you're left with
$$
\lim_{n\to\infty}n\cos\dfrac{\pi n}{2n+1}=
\lim_{n\to\infty}n\sin\left(\frac{\pi}{2}-\dfrac{\pi n}{2n+1}\right)=
\lim_{n\to\infty}\frac{\sin\dfrac{\pi}{2(2n+1)}}{\dfrac{\pi}{2(2n+1)}}
\frac{\pi n}{2(2n+1)}=
\frac{\pi}{4}
$$
Now just put all pieces together and the limit is $\dfrac{4}{\pi}$.
A: Another way is based on the asymptotics of the numerator and denominator $$\tan \left(\frac{\pi  n}{2 n+1}\right)=\frac{4 n}{\pi }+\frac{2}{\pi }-\frac{\pi }{12 n}+\frac{\pi }{24
   n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ $$\sqrt[3]{n^3+2 n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ Making the long division,then $$\dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}=\frac{4}{\pi }+\frac{2}{\pi  n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
