Limit of $\frac{2^n}{n^{\sqrt{n}}}$ $$\lim_{n\to\infty}\frac{2^n}{n^{\sqrt{n}}}$$
The way I see it is that denominator is super-exponential and the numerator is only exponential, so this must be $0$ but wolfram is saying it is infinite.
How do I evaluate this limit?
 A: You need  some (easy) asymptotic analysis to obtain the limit.
As already noted
$$\frac{2^n}{n^{\sqrt{n}}}=\mathrm e^{n\ln 2-\sqrt{n}\ln n}=\mathrm e^{n\ln 2\bigl(1-\tfrac{\ln n}{\sqrt n\ln 2}\bigr)}.$$
Now it is standard that 
$$\ln n=o(\sqrt n)\enspace\text{ –  in other words },\; \lim_{n\to\infty} \frac{\ln n}{\sqrt n}=0.$$
Therefore, the second factor in the exponent tends to $1$, and consequently the exponent  itself tends to $+\infty$.
A: Hint:
You can rewrite it as follows:
$$\lim_{n \to \infty} \exp{(n \ln{2}-\sqrt{n} \ln{n})}$$
Can you take it from here?
Edit: Now you only need to notice, that $n \ln{2}$ grows faster than $\sqrt{n} \ln{n}$.
A: By substituting $n=2^{t}$, we obtain $${2^n\over n^{\sqrt n}}={2^{2^t}\over 2^{t\cdot 2^{t\over 2}}}=2^{2^t-t\cdot 2^{t\over 2}}$$which tends to $\infty$.
A: HINT.-(A way to see it) The derivative of $f(x)=\dfrac{2^x}{x^{\sqrt{x}}}$ is equal to $\dfrac{2^{x-1}}{x^{\sqrt x-0.5}}(\sqrt x\ln4-2-\ln x)$ and it has only one zero at $x_0\approx 9.319913$. It is not difficult to see that $f$ has a minimum at $x_0$ then that  for $x\gt x_0$   the function is strictly increasing. You can deduce this way that your limit is not $0$ and it is according with Wolfram says.
A: Let us suppose, $a^2<n<(a+1)^2$
It means $2^{\sqrt n}>2^a>(a+1)^2>n$ for $ n>5$, 
So now we have $\frac{2^{\sqrt n}}{n}>1$ hence the limit is infinity. 
again if,  $n=a^2$ then we have $ 2^a>a^2$ again the fraction is larger than $1$ hence $c_n^{\sqrt n}$ tends to $\infty$, 
With $c_n>1$. also $c_{n+1}>c_n$,  where $c_n=\frac {2^{\sqrt n}}{n}$.
Note :  the function $\frac {2^{\sqrt n}}{n}$ is increasing. 
