Find the limit $\lim_{n\rightarrow\infty}\frac{(100n)!}{(99n)!(100n)^n}$ 
Find the limit 
  $$\lim_{n\rightarrow\infty}\frac{(100n)!}{(99n)!(100n)^n}$$

How to prove that this limit is $0$? I've tried to find it's upper estimate that tends to zero, but did not find something better than with limit $1$.
 A: Let $a_n=\frac{(100n)!}{(99n)!(100n)^n}$ and evaluate the limit of the ratio
$$\begin{align}\frac{a_{n+1}}{a_n}&=\frac{(100(n+1))!}{(99(n+1))!(100(n+1))^{n+1}}\cdot
\frac{(99n)!(100n)^n}{(100n)!}\\
&=\frac{(100n+99)\cdots (100n+1)}{(99n+99)\cdots (99n+1)(1+\frac{1}{n})^n}\to
\frac{(1+\frac{1}{99})^{99}}{e}<1.\end{align}$$
Then apply the ratio test for sequences (see Proof attempt to the ratio test for sequences).
A: Note that:
$$
\frac{100n!}{99n!(100n)^n} = \frac{(99n+1)(99n+2)\ldots (99n+n-1)(100n)}{(100n)^n} = \prod_{k=1}^{n} \frac{99n+k}{100n} \leq \Big(\frac{99n+n/2}{100n}\Big)^{n/2}
$$
where the inequality at the end comes from using 1 as an upper estimate for the last $n/2$ terms and $\Big(\frac{99n+n/2}{100n}\Big)$ as an upper estimate for the first $n/2$ terms. By canceling the $n$'s inside this is just equal to $(99.5/100)^{n/2}$ which clearly goes to 0. 
What I like about this approach is that it doesn't rely on knowing/ proving other results about famous limits.
A: You could use Stirling approximation for $ x \to \infty$:
$$x!\approx \sqrt{2\pi}\frac{x^{x+\frac{1}{2}}}{e^x}$$
So the limit becomes:
$$\lim_{x \to \infty} \frac{\sqrt{2\pi}\frac{(100x)^{100x+\frac{1}{2}}}{e^x}}{\sqrt{2\pi}\frac{(99x)^{100x+\frac{1}{2}}}{e^x}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{100x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}(100x)^x}=\lim_{x \to \infty} \frac{(100x)^{99x+\frac{1}{2}}}{(99x)^{100x+\frac{1}{2}}}= $$
$$=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(100x)^{99x}}{(99x)^{100x}}=\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x} $$
For infinites hierarchy $ x^x>>a^x $
:
$$\frac{10}{\sqrt{99}}\lim_{x \to \infty} \frac{(\frac{100}{99})^{99x}}{(99x)^x}=0 $$
