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.

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.
