Find $k$ such that $\lim_{n\to\infty}\tan n/n^k=0$. Find $k\in\mathbb{R}$ such that $$\lim_{n\to\infty}\frac{\tan n}{n^k}=0.$$


Does such a $k$ exist? And is it possible to find all such $k$?

Edit: It is $\lim_{n\to\infty}\tan n/n^k=0$, the limit of the sequence, instead of $\lim_{n\to\infty}\tan x/x^k=0$. I was wrong. I'm really sorry for the trouble I caused...
I expect you really mean $$\lim_{n\to\infty}\frac{\tan n}{n^k}=0.$$
Here the limit is over integers, not reals. The difficulty here is that
an integer $n$ is close to an odd multiple of $\pi/2$. If that is so
then $|n-m\pi/2|$ is small and then $|\tan n|\approx|n-m\pi/2|^{-1}$.
If we had a theorem stating that $|n-m\pi/2|&gt;Cn^{-r}$ then for $k&gt;r$
your limit would be zero. This leads to the "irrationality measure of $\pi$". Theorems of this nature are known but difficult. I am not up to speed on recent developments, but you might want to start with
Zudilin's survey.
An easy proof of this is given by an intuitive reasoning using the definition of a limit. We write: 

$$\lim_{x\rightarrow\infty}f(x)=L \text{ iff } \forall\epsilon&gt;0, \exists N \text{ s.t. } \forall x\geq N, |L-f(x)|&lt;\epsilon$$

In this case: 

$$L=0, f(x)=\frac{\tan(x)}{x^k}$$

And let's have $\epsilon=1$, an easy number. Now, all we have to do is prove that there is no $N$ s.t. for all $x\geq N$, 

$$\left|-\frac{\tan(x)}{x^k}\right|&lt;1$$

Remember, $1$ is our $\epsilon$. 

Proof by contradiction. Suppose $N$ exists. Then, $N$ cannot be represented by $\pi/2+k_1\pi$, $k_1 \in \mathbb{N}$, because that our above statement is undefined and therefore not less than $1$. In fact, if our function in the absolute value sign (let's call it $A(x)$ for convenience), is undefined at any point $c$ larger than $N$, it isn't smaller than $1$. 

Since $N$ cannot be represented by $\pi/2+k_1\pi$ for all $k_1$, it can be said that: 

$$\frac{\pi}2+m\pi &lt; N &lt; \frac{\pi}{2}+(m+1)\pi, m\in\mathbb{N}$$

Let's have $v=\frac{\pi}{2}+(m+1)\pi$

Remember two things: 


$v&gt;N$
$A(x)$ is undefined at $v$, because $A(x)$ is undefined at all points $\pi/2+k_1\pi$ for all $k_1 \in \mathbb{N}$


This is a contradiction.

So, there does exist an $\epsilon$ such that no $N$ exists to fill the requirement, no matter what $k$ is. 

This means that it is not the case that for all epsilon that there exists an $N$ to fufill the requirement. 

Q.E.D

This is really rigorous, but that, in my opinion, is the best kind of proof. It's like responding to the YouTube comments with your pretentious a****** vocabulary. Can I swear on StackExchange? I should check. Bleeping it out anyway, I guess. 

Have a nice day, whoever's reading this. 

OH NO, THE STACK EXCHANGE POLICE ARE BREAKING DOWN THE DOOR! THEY'RE CO