calculating a limit of sequence with tan

Question

Calculating a limit

$$\lim_{n\longrightarrow +\infty} \dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}$$

thanks.

Answer 1

1. Change the variable $n \to \frac 1n$ $$ L = \lim_{n \to 0^+} \frac {\tan \left ( \frac {\frac \pi n}{\frac 2n+1}\right )}{\sqrt[3]{\frac 1{n^3}+\frac 2n - 1}} = \lim_{n \to 0^+} \frac {n \tan \left ( \frac \pi{2+n} \right )}{\sqrt[3]{1+2n^2-n^3}} = \lim_{n \to 0^+} \left [ n \tan \left( \frac \pi{2+n} \right ) \right ] $$
2. Use $\tan \alpha = \frac 1{\tan \left(\frac \pi 2 - \alpha \right )}$ $$ L = \lim_{n \to 0^+} \frac n{\tan \left ( \frac \pi 2 - \frac \pi{2+n} \right )} = \lim_{n \to 0^+} \frac n{\tan \left( \frac {\pi n}{4+2n}\right )} = \lim_{n \to 0^+} \frac n{\frac {\pi n}{4+2n}} = \lim_{n \to 0^+} \frac {4+2n}\pi = \frac 4\pi $$
3. Check - WA.

Answer 2

Hint : $cos(\frac{\pi}{2}+x)$ is equivalent to $-x$ when $ x \to 0$ and $\frac{\pi n}{2n+1}=\frac{\pi}{2}(1-\frac{1}{2n+1} )$

Answer 3

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

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

Answer 4

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 $\dfrac{4}{\pi}$.

Answer 5

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.