<s>[INST] 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.
 [/INST] 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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\tan\left(\frac{\pi n}{2n+1}\right)}{\left(n^3+2n-1\right)^{1/3}}=\frac{4}{\pi}}$$
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