I have been having trouble with questions with factorials in Squeeze Theorem.
This is the questions that I am struggling with:
$\lim_{x\to \infty} {x^x\over(2x)!}$
What I have done so far:
Lower Limit = $0$ because the lowest value that the function can be is when $n=0$.
I tried to workout the upper limit as follows:
${x^x \over (2x)!} = {x \over 2x} \times {x \over (2x-1)} \times {x \over (2x-2)} ... \times {x \over (2x - (x-1))}$
$= {x^x \over (2x)!} = {x \over 2x} \times {x \over (2x-1)} \times {x \over (2x-2)} ... \times {x \over (x+1)}$
However, I don't know how to proceed from this point on. Any hints on tackling this problem would be much appreciated. Any tips on solving factorials with squeeze theorem in general would also be be helpful.
I will use $n$ instead of $x$. Note that if $n\ge 1$ then $$\frac{n^n}{(2n)!}=\frac{1}{n!}\left (\frac{n}{n+1}\cdot\frac{n}{n+2}\cdot\frac{n}{n+3}\cdots \frac{n}{2n} \right).$$ But it is clear that $$\frac{n}{n+1}\cdot\frac{n}{n+2}\cdot\frac{n}{n+3}\cdots \frac{n}{2n} \lt 1.$$ Thus if $n\ge 1$ then $$0\lt \frac{n^n}{(2n)!}\lt \frac{1}{n!}.$$ As $n\to\infty$, the quantity $\frac{1}{n!}\to 0$. It follows by Squeezing that $$\lim_{n\to\infty}\frac{n^n}{(2n)!}=0.$$
Remark: The question asked for a general procedure. That I cannot provide. There are recurring themes, and after doing a number of problems one gets accustomed to some of them.