Find limit of $\lim\limits_{x \to\infty}{\left({{(x!)^2}\over{(2x)!}}\right)}$ I'm practising solving some limits and, currently, I'm trying to solve $\lim\limits_{x\to\infty}{\left({{(x!)^2}\over{(2x)!}}\right)}$.
What I have done:


*

*I have attempted to simplify the fraction until I've reached an easier one to solve, however, I'm currently stuck at the following:


$$
\lim_{x→\infty}{\left({{(x!)^2}\over{(2x)!}}\right)}=
\lim_{x→\infty}{\left({{(\prod_{i=1}^{x}i)^2}\over{\prod_{i=1}^{2x}i}}\right)}=
\lim_{x→\infty}{\left({
   {
      {\prod_{i=1}^{x}i}\cdot{\prod_{i=1}^{x}i}
   }\over{
   {
      {\prod_{i=1}^{x}}i}\cdot{\prod_{i=x+1}^{2x}i}
   }
}\right)}=
\lim_{x→\infty}{\left({
   {\prod_{i=1}^{x}i}\over{
   {\prod_{i=x+1}^{2x}i}}
}\right)}.
$$


*

*Instinctively, I can see that the limit is equal to $0$, since the numerator is always less than the denominator, thus approaching infinity slower as $x→\infty$.


Question:


*

*How can I continue solving the above limit w/o resorting to instinct to determine it equals $0$ ?

*If the above solution can't go any further, is there a better way to approach this problem?


Stirling formula may be difficult to remember, but the simpler one below is extremely useful and allows you to solve most asymptotic results with factorials: 
Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$
You get $\quad\dfrac{n^{2n}}{9^n}< (n!)^2 < \dfrac{n^{2n}}{4^n}$
And also $\quad\dfrac{4^nn^{2n}}{9^n}< (2n)! < \dfrac{4^nn^{2n}}{4^n}$
So by dividing positive quantities $0<\dfrac{(n!)^2}{(2n)!}<\dfrac{n^{2n}9^n}{4^n4^nn^{2n}}=\left(\dfrac 9{16}\right)^n\to 0$

You can also notice that in this case $\dfrac{(2n)!}{(n!)^2}=C_{2n}^n\ge C_{2n}^1\ge 2n\to\infty$ so its inverse is going to zero.
