Proving $\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$ For any real $a>0$:
$$\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$$
I got the hint to use Stirling's approximation but I cant seem to solve it. Any tips would be very appreciated.
 A: Let $ a,x\in\left(1,+\infty\right) $, we have the following : $$ x^{a}\int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}y}=\int_{0}^{x}{y^{a-1}\left(1-\frac{y}{x}\right)^{x-1}\,\mathrm{d}y}=\int_{0}^{+\infty}{y^{a-1}f_{x}\left(y\right)\mathrm{d}y} $$
Were $ f_{x} $ is the function defined on $ \mathbb{R}_{+} $ as follows : $$ \left\lbrace\begin{matrix}f_{x}\left(y\right)=\left(1-\frac{y}{x}\right)^{x-1},\ \ \ \ \ \ \ \textrm{If }y\leq x\\ f_{x}\left(y\right)=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,\ \ \ \ \ \ \ \textrm{If }y\geq x \end{matrix}\right. $$
Using the fact that $ \left(\forall y\in\mathbb{R}_{+}\right),\ f_{x}\left(y\right)\leq\mathrm{e}^{-y}=f\left(y\right)$, and that $ f_{x}\left(y\right)\underset{x\to +\infty}{\longrightarrow}f\left(y\right) $, by using the dominated convergence theorem, we have the following : $$ \int_{0}^{+\infty}{y^{a-1}f_{x}\left(y\right)\mathrm{d}y}\underset{x\to +\infty}{\longrightarrow}\int_{0}^{+\infty}{y^{a-1}f\left(y\right)\mathrm{d}y} $$
In other words : $$ x^{a}\int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}x}\underset{x\to +\infty}{\longrightarrow}\int_{0}^{+\infty}{y^{a-1}\,\mathrm{e}^{-y}\,\mathrm{d}y} $$
But $ \int_{0}^{1}{y^{a-1}\left(1-y\right)^{x-1}\,\mathrm{d}x}=\beta\left(a,x\right)=\frac{\Gamma\left(a\right)\Gamma\left(x\right)}{\Gamma\left(a+x\right)} $, and $ \int_{0}^{+\infty}{y^{a-1}\,\mathrm{e}^{-y}\,\mathrm{d}y}=\Gamma\left(a\right) $, thus : $$\fbox{$\begin{array}{rcl} \displaystyle\frac{\Gamma\left(x\right)}{\Gamma\left(a+x\right)}\underset{x\to +\infty}{\sim}\frac{1}{x^{a}} \end{array}$}$$
It's another way of saying that $ \lim\limits_{x\to +\infty}{\frac{\Gamma\left(a+x\right)}{x^{a}\Gamma\left(x\right)}}=1 $.
A: I just solved it and for anyone interested the answer 
$ \frac{\Gamma(x+a)}{x^a\Gamma(x)}=\frac{\sqrt{2\pi(x+a)}(\frac{x+a}{e})^{x+a}}{\sqrt{2\pi(x)}(\frac{x}{e})^{x}}=
 \frac{\sqrt{2\pi }\sqrt{x+ a}(x+a)^{x+a}e^x}{\sqrt{2\pi }\sqrt{x }(x)^{x}e^{x+a}}=
(\frac{x+a}{x})^{\frac{1}{2}ax}\cdot\frac{1}{e^a}  \to \frac{e^a}{e^a}=1 $
