This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$,
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts.
$\endgroup$ 63 Answers
$\begingroup$Using the addition angle formula for the tangent function we can write
$$\tan^n\left(\frac{\pi}{4}+\frac1n\right)=\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n$$
Note that $\tan(x)=x+O(x^3)$. Then, we have
$$\begin{align} \left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n&=\left(\frac{1+\frac1n +O\left(\frac1{n^3}\right)}{1-\frac1n +O\left(\frac1{n^3}\right)}\right)^n\\\\ &=\left(\frac{1+\frac1n }{1-\frac1n }\right)^n\left(\frac{1+\frac{O\left(\frac1{n^3}\right)}{1+\frac1n}}{1 +\frac{O\left(\frac1{n^3}\right)}{1-\frac1n }}\right)^n\\\\ &\to \frac {e}{e^{-1}}\,\frac{1}{1}\,\,\text{as}\,\,n\to \infty\\\\ &=e^2 \end{align}$$
And we are done!
$\endgroup$ 10 $\begingroup$If the only standard limit available is $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ then we are out of luck here. It is better to assume the following standard limits $$\lim_{x \to 0}\frac{\tan x}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}$$ and then we can easily evaluate the desired limit. We have \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\tan^{n}\left(\frac{\pi}{4} + \frac{1}{n}\right)\right\}\notag\\ &= \log\left\{\lim_{n \to \infty}\left(\frac{1 + \tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\ &= \log\left\{\lim_{n \to \infty}\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\ &= \lim_{n \to \infty}\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\text{ (via continuity of log)}\notag\\ &= \lim_{n \to \infty}n\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)\notag\\ &= \lim_{n \to \infty}n\cdot\dfrac{2\tan(1/n)}{1 - \tan(1/n)}\cdot\dfrac{\log\left(1 + \dfrac{2\tan(1/n)}{1 - \tan(1/n)}\right)}{\dfrac{2\tan(1/n)}{1 - \tan(1/n)}}\notag\\ &= 2\lim_{n \to \infty}\dfrac{\tan(1/n)}{1/n}\cdot\dfrac{1}{1 - \tan(1/n)}\cdot\lim_{x \to 0}\dfrac{\log\left(1 + x\right)}{x}\notag\\ &= 2\cdot 1\cdot 1\cdot 1\notag\\ &= 2\notag \end{align} Hence $L = e^{2}$.
$\endgroup$ $\begingroup$Solution
Applying $\textbf{L'Hospital's Rule}$, we have $$\begin{align*}\lim_{x \to +\infty} \left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]&=\lim_{x \to +\infty}\dfrac{\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&=\lim_{x \to +\infty}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\cos \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \sin\left(\dfrac{\pi}{2}+\dfrac{2}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \cos\left(\dfrac{2}{x}\right)}\\&=2. \end{align*}$$ Hence, $$\begin{align*}\lim\limits_{n\to \infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right)&=\lim\limits_{x\to +\infty}\tan^x\left(\frac{\pi}{4}+\frac{1}{x}\right)\\&=\lim\limits_{x\to +\infty} \exp\left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\exp\left[\lim_{x \to \infty}x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\mathbb{e}^2.\end{align*}$$
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