Find $a$ such that the limit is zero

Question

Problem :

Let $x>0$ then define :

$$f(x)=\left(\left(\frac{1}{x}\right)!\left(x!\right)\right)^{\frac{1}{x+\frac{1}{x}}}$$

Then find $a$ such that :

$$\lim_{x\to\infty}f(x)-\frac{1}{2}\left(\frac{1}{x^{a}}+x^{a}\right)=0$$

First I don't know if $a$ exists so it should be a first step .If $a$ exists my intuition says $0<a<1$.

For the motivation I see the inequality in my book for $x>0$ :

$$\left(\frac{1}{x}\right)!\left(x!\right)\geq 1$$

So I was wondering if we can find a Stirling formula for this product .

How to find $a$ if it exists? Does it admits a closed form or an infinite series ?

Answer 1

Using Stirling approximation $$\log\Big[\frac{1}{x}!\,\, x!\Big]=x (\log (x)-1)+\frac{1}{2} \log (2 \pi x)+\frac{1-12\gamma }{12x}+\frac{\pi ^2}{12 x^2}+O\left(\frac{1}{x^3}\right)$$

Such an $a$ exists; it is just smaller than $1$.

$$g(a)=\lim_{x\to\infty}\Bigg[\left(\frac{1}{x}!\,\, x!\right)^{\frac{x}{x^2+1}}-\frac{1}{2}\left(x^a+\frac{1}{x^{a}}\right)\Bigg]$$

$g(1)$ is always negative since, by Taylor $$g(1)=-\left(\frac{1}{2}-\frac{1}{e}\right) x+\frac{\log (2 \pi x)}{2 e}+O\left(\frac{1}{x}\right)\quad \to \quad -\infty$$ $$g\left(\frac{999}{1000}\right)=+\infty$$