Evaluating $\lim_{n \to \infty}\tan(n\pi/(2n+1))^{1/n}$

Question

My input

$\lim \limits_{n \to \infty}\tan\bigg(\dfrac{n\pi}{2n+1}\bigg)^{\frac{1}{n}}$

$\lim \limits_{n \to \infty}\Bigg(\dfrac{1}{\frac{2n+1}{n\pi}}\Bigg)^\frac{1}{n}\cdot\underbrace{\dfrac{\tan\Bigg(\dfrac{1}{\frac{2n+1}{n\pi}}\Bigg)^{\frac{1}{n}}}{\Bigg(\dfrac{1}{\frac{2n+1}{n\pi}}\Bigg)^{\frac{1}{n}}}}_{\text{=1}}$

$\lim \limits_{n \to \infty}\bigg(\dfrac{n\pi}{2n+1}\bigg)^{\frac{1}{n}}$

I will proceed further If I know that I am on correct path. Someone help me here.

Answer 1

We have that

$$\bigg(\dfrac{n\pi}{2n+1}\bigg)=\frac{\pi}2-\dfrac{\pi}{4n+2}\implies \tan \bigg(\dfrac{n\pi}{2n+1}\bigg)=\frac1{\tan \bigg(\dfrac{\pi}{4n+2}\bigg)}\sim \dfrac{4n}{\pi}$$

and therefore

$$\left[\tan \bigg(\dfrac{n\pi}{2n+1}\bigg)\right]^{\frac{1}{n}}\sim \left(\dfrac{4n}{\pi}\right)^{\frac{1}{n}}\to 1$$

Answer 2

In case you meant \begin{equation} \tan \Bigg( \Big(\frac{n\pi}{2n + 1}\Big)^{\frac{1}{n}}\Bigg) \end{equation} Let $L = \lim \limits_{n \to \infty}\bigg(\dfrac{n\pi}{2n+1}\bigg)^{\frac{1}{n}}$ \begin{equation} \log L = \lim_{n \rightarrow \infty} \frac{1}{n} \log \frac{n\pi}{2n+1} \end{equation} As $n$ goes to infinity, $\frac{n\pi}{2n+1} $ goes to $\frac{\pi}{2}$ so \begin{equation} \log L = \lim_{n \rightarrow \infty} \frac{\log \frac{\pi}{2}}{n} = 0 \end{equation} So \begin{equation} L = e^{0} = 1 \end{equation} So the limits above $$\lim \limits_{n \to \infty}\tan\bigg(\dfrac{n\pi}{2n+1}\bigg)^{\frac{1}{n}} = \tan 1$$

In case you meant \begin{equation} \Big( \tan \frac{n\pi}{2n + 1} \Big)^{\frac{1}{n}} \end{equation} Let \begin{equation} L = \lim_{n \rightarrow \infty} \log \Big( \tan \frac{n\pi}{2n + 1} \Big)^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \frac{1}{n} \log \tan \frac{n\pi}{2n + 1} = \lim_{n \rightarrow \infty} \frac{\log \tan \frac{n\pi}{2n + 1}}{n} = \lim_{n \rightarrow \infty} \frac{\infty}{\infty} \end{equation} Applying L'Hopital to the above, we get \begin{equation} L = \lim_{n \rightarrow \infty} \frac{ \big( \log \tan \frac{n\pi}{2n + 1}\big)^{'}}{n^{'}} = \frac{\frac{1}{(2n+1)^2}[1 + \cos^2(\frac{n\pi}{2n + 1})]}{\tan \frac{n\pi}{2n + 1}} = \frac{[1 + \cos^2(\frac{n\pi}{2n + 1})]}{(2n+1)^2\tan \frac{n\pi}{2n + 1}} \end{equation} The numerator is bounded and denominator goes to infinity, so \begin{equation} L = 0 \end{equation} and \begin{equation} \Big( \tan \frac{n\pi}{2n + 1} \Big)^{\frac{1}{n}} \rightarrow e^0 = 1 \end{equation}

Answer 3

Use the fact that $\tan=\frac{\sin}\cos$: $$ \lim_{n\to\infty}\left(\frac{\sin\big(\frac\pi2(1-\frac1{n})\big)}{\cos\big(\frac\pi2(1-\frac1{n})\big)}\right)^{1/n} $$ Now use the symmetry $\cos(\theta)=\sin(\frac\pi2-\theta)$ $$ \lim_{n\to\infty}\left(\frac{\cos\big(\frac\pi{2n}\big)}{\sin\big(\frac\pi{2n}\big)}\right)^{1/n} $$ Finally, use the small order terms of the Taylor series for $\sin,\cos$: $$ \lim_{n\to\infty}\left(\frac{1+O(1/n^2)}{\frac\pi{2n}+O(1/n^3)}\right)^{1/n}=\lim_{n\to\infty}(n+O(1/n^2))^{1/n} $$ Since $\lim_{n\to\infty}n^{1/n}=1$, the same is true for $\lim_{n\to\infty}(n+O(1/n^2))^{1/n}$.