How to calculate $\lim_{n\to+\infty}\frac{\Gamma(n)}{\Gamma\left(n-x\sqrt n\right)}\frac{1}{n^{x\sqrt n}}\,$ given that $x\in\mathbb R$?

Question

How to calculate $\displaystyle\lim_{n\to+\infty}\frac{\Gamma(n)}{\Gamma\left(n-x\sqrt n\right)}\frac{1}{n^{x\sqrt n}}\,$ given that $x\in\mathbb R$?

I know that it should be $e^{-\frac{x^2}{2}}$, but I don't know the steps.

Answer 1

Let us consider $$A_n=\frac{\Gamma(n)}{\Gamma\left(n-x\sqrt n\right)}\,\frac{1}{n^{x\sqrt n}}$$ Take logarithms $$\log(A_n)=\log\big(\Gamma(n)\big)-\log\big(\Gamma\left(n-x\sqrt n\right)\big)-x \sqrt n \log(n)$$ Now, use Stirling approximation $$\log\big(\Gamma(z)\big) \sim (z-\frac 12)\log( z) - z + \frac12\log(2\pi) $$ Applying it, we then have $$\log(A_n)\sim -x\sqrt{n} +\left(-x\sqrt{n} +n-\frac{1}{2}\right) \log (n)+\left(x\sqrt{n} -n+\frac{1}{2}\right) \log \left(n-x\sqrt{n} \right)$$ Now, rewrite $$\log \left(n-x\sqrt{n}\right)=\log(n)+\log(1-\frac x {\sqrt n})\sim \log(n)-\frac x {\sqrt n}-\frac{x^2}{2n}$$ Replacing in the previous expression, after simplifications, $$\log(A_n)\sim -\frac{x}{2 \sqrt{n}}-\left(\frac{1}{4 n}+\frac{1}{2}\right) x^2-\frac{x^3}{2 \sqrt{n}}$$ So, for infinitely large values of $n$, $\log(A_n)\sim -\frac{x^2}2$