Evaluating $\lim_{n\to\infty} \frac{a^n}{n^\sqrt n}$ 
Evaluate the limit $$\lim_{n\to\infty} \frac{a^n}{n^\sqrt n}$$ where $a\in\mathbb{R}$

I've tried several approaches but failed (I guess my limit evaluation ability is a bit rusty). 
I'd be glad for guidance!
 A: If $a> 0$:
Compute the $\log$:
$$
\log a_n = n \log a - \sqrt n\log n 
$$
Now two cases:


*

*$a\le 1$: $\log a_n\to-\infty$ and  $a_n\to 0$

*$a>1$: you can prove that $$\log a_n > \frac 12 n\log a $$ for $n$ big enough. Hence
$a_n\to \infty$.


If $a=0$: 
this is $0$.
If $a<0$:
Use the results for $a>0$ with $|a_n| $ instead of $a_n$.
A: Hint :
$\log a_n = n \log a - \sqrt n\log n=n \cdot(\log(a)-\frac{\log(n)}{\sqrt{n}})$ and $\frac{\log(n)}{\sqrt{n}}$ is well known to have a limit of $0$.
A: == If $\;a=0\;$ the limit is zero
== If $\;-1<a<0\;$ then both $$\;\begin{cases}a^n\xrightarrow[n\to\infty]{}0\\{}\\\frac1{n^{\sqrt n}}\xrightarrow[n\to\infty]{}0\end{cases}\;\implies\frac{a^n}{n^{\sqrt n}}\xrightarrow[n\to\infty]{}0\;$$
== If $\;a<-1\;$ then $\;\{a^n\}_{n\in\Bbb N}\;$ swings between positive and negative values, so our only chance for convergence is if the denominator of the sequence converges to zero faster than the numerator, so taking even $\;n$' s to make things positive, we have a case already covered by other answers.
