How to prove that the limit doesn't exist I suspect that the limit of the following sequence doesn't exist as $n$ approaches infinity.
$$x_n= n^{(1/2-\sin n)}$$
Wolfram alpha is unable to compute this limit. However, I it seems intuitive by looking at its plot.
How can I prove this result?
 A: The limit does not exist because $$\liminf_{n\to \infty} x_n=0 \neq +\infty= \limsup_{n\to \infty} x_n$$ To see this you need to take a subsequence $n_k$ with $\sin n_k<1/2$ for all $n_k \in \mathbb N$ and another $n_m$ with $\sin n_m>1/2$ for all $n_m \in \mathbb N$. (Due to periodicity of $\sin$ there are infinitely many of both, so you can find such sub-sequences). 
A: The values of $\sin n$ are the values of the $y$ coordinates of the points on the unit circle that you land on when marching counterclockwise starting from $(1,0)$ in steps of arc length $1$.  Now since the arcs 
$$(2k\pi+{\pi\over2}-{\pi\over6},2k\pi+{\pi\over2}+{\pi\over6})\quad\text{and}\quad(2k\pi-{\pi\over2}-{\pi\over6},2k\pi-{\pi\over2}+{\pi\over6})$$ 
are all of arc length ${\pi/3}\gt1$, the march will land in each of them.  
For the first batch, $\sin n\gt\sin{\pi\over3}={\sqrt3\over2}\gt{1\over2}$, and for the second, $\sin n\lt-\sin{\pi\over3}\lt-{\sqrt3\over2}\lt-{1\over2}$.  Thus infinitely often the sequence $n^{(1/2-\sin n)}$ takes values less than $n^{(1-\sqrt3)/2}$, which is less than $1$ for $n\ge2$, and infinitely often it takes values greater than $n^{(1+\sqrt3)/2}$, which is greater than $2$ for $n\ge2$.  Hence the sequence does not have a limit as $n\to\infty$.
Remark:  It's not hard to show that the aforementioned subsequences tend, in fact, to $0$ and $\infty$.  It might be of interest to ask if there are subsequences that tend to anything in between -- i.e., are there "Goldilocks" integers $n$ that that $\sin n$ is close enough to $1/2$ to make $n^{(1/2-\sin n)}$ neither too big nor too small?
