calculus - Find limit of $limlimits_{x toinfty}{left({{(x!)^2}over{(2x)!}}right)}$

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?

Answer

Continuing from what you have mentioned,

$$0 \le \lim_{x\to\infty}{\left({
{\prod_{i=1}^{x}i}\over{
{\prod_{i=x+1}^{2x}i}}
}\right)} = \lim_{x\to\infty}\prod_{i=1}^{x}\frac{i}{i+x} \le \lim_{x\to\infty}\prod_{i=1}^{x}\frac{x}{x+x} = \lim_{x\to\infty}\frac{1}{2^x}=0.$$