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...
 A: 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 $|n-m\pi/2|>Cn^{-r}$ then for $k>r$
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.
A: An easy proof of this is given by an intuitive reasoning using the definition of a limit. We write: 
$$\lim_{x\rightarrow\infty}f(x)=L \text{ iff } \forall\epsilon>0, \exists N \text{ s.t. } \forall x\geq N, |L-f(x)|<\epsilon$$
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 $x\geq N$, 
$$\left|-\frac{\tan(x)}{x^k}\right|<1$$
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: 
$$\frac{\pi}2+m\pi < N < \frac{\pi}{2}+(m+1)\pi, m\in\mathbb{N}$$
Let's have $v=\frac{\pi}{2}+(m+1)\pi$
Remember two things: 


*

*$v>N$

*$A(x)$ 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
