Finding the limit of a function $n^{n}/e^{n^{3/2}}$ What is the limit
$$\lim_{n\to\infty}\frac{n^{n}}{e^{n^{3/2}}}?$$
 A: Observe that $n^n = e^{n\ln n}$. So the limit becomes
$$\lim_{n \to \infty} \frac{e^{n\ln n}}{e^{n^{3/2}}}.$$
Without applying l'Hospital's Rule, which one do you think grows faster: $n\ln n$ or $n^{3/2}$? This is a shortcut (l'Hospital's rule in disguise).
A: We will be using this result
Theorem: If ${a_n}$ be a sequence such that $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}= a\,,$ then
1) if $|a|<1$, then $\lim_{n\to \infty}a_n =0 \,,$ 
2) if $ a>1$, then $\lim_{n\to \infty}|a_n| =\infty \,.$
Another fact, we need to achieve our task is 
$$ (n+1)^{\frac{3}{2}} = n^{\frac{3}{2}} + \frac{3}{2}\sqrt{n} + O(\frac{1}{\sqrt{n}})\,, $$
which can be derived by the binomial theorem. Now, let $ a_n = \frac{n^{n}}{e^{n^{3/2}}}\, $ then
$$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{n^{n+1} e^{n^{3/2}}}{ e^{(n+1)^{3/2} }n^n } = \lim_{n\to\infty} \frac{n e^{n^{3/2}}}{e^{n^{3/2}+\frac{3}{2}\sqrt{n}+O(1/\sqrt{n})}}= \lim_{n\to\infty} \frac{n}{e^{\frac{3}{2}\sqrt{n}+O(1/\sqrt{n})}}=0\,.$$
Since $ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0 \,,$ then by part $(a)$ of the theorem $\lim_{n\to\infty} a_n = 0. $  
