Calculating $\lim\limits_{j \to \infty} \frac{j^{j/2}}{j!}$ \begin{align*}
 & \lim\limits_{j \to \infty}{j^{\,j/2} \over j!} \\
\end{align*}
This problem is from a real analysis textbook in the chapter on the natural log and properties of exponents. I'm struggling with how to approach this. I don't think you would use L'Hopital's rule.
 A: The ratio test is a more elementary tool:
$$\frac{(j+1)^\tfrac{j+1}2}{(j+1)!}\,\frac{j!}{j^{\tfrac j2}}=\frac{(j+1)^{\tfrac{j}2}\sqrt{j+1}}{(j+1)j^{\tfrac j2}}=\underbrace{\sqrt{\Bigl(1+\frac1j\Bigr)^j}}_{\begin{array}{c}\downarrow\\\sqrt{\mathrm e}\end{array}}\,\frac1{\sqrt{j+1}}\to 0$$
A: Without invoking Stirling's approximation, the Hermite-Hadamard inequality, the $\log$-convexity of $\Gamma$ or the trapezoid rule (all viable approaches), one may simply notice that by defining
$$ a_n = \frac{n^n}{n!^2} $$
we have 
$$ \frac{a_{n+1}}{a_n} = \frac{1}{n+1}\underbrace{\left(1+\frac{1}{n}\right)^n}_{\text{bounded}} \to 0 $$
as $n\to +\infty$, hence your limit is zero as well.
A: Let $L$ be the limit. Then, it is :
$$L = \lim_{n \to \infty} \frac{n^{n/2}}{n!} \Leftrightarrow \ln L = \lim_{n \to \infty} \left(\ln n^{n/2}-\ln n! \right)$$
$$\Leftrightarrow$$
$$\ln L =\lim_{n \to \infty} \left(\frac{n}{2}\ln n - \ln n!\right)$$
Now we will use the fact 
$$\ln n! = n \ln n - n + \mathcal{O}\left(\ln n\right)$$
which is called Stirling's Approximation (actually it is a consquence of the original formula, more information can be found in the link and credits to Jack D'Aurizio for mentioning it as well).
$$\begin{align*}\ln L &=\lim_{n \to \infty} \left(\frac{n}{2}\ln n - n \ln n + n - \mathcal{O}\left(\ln n\right)\right)\\&=\lim_{n \to \infty} \left(n-\frac{n}{2}\ln n\; -\mathcal{O}\left(\ln n\right)\right) \end{align*}$$
But $\mathcal{O}\left(\ln n\right) \to 0$ and you should be able to finish now.
