Q:

Ratio of 2 Gammas, approximation with power

Find all value of $\alpha$ such that $\lim\limits_{x\rightarrow +\infty}\left(\frac{\Gamma(x+\alpha)}{\Gamma(x)}-x^{\alpha}\right)=0$. (note: $\alpha$ is a constant with respect to $x$)
By restricting to just the integer case ($\Gamma$ reduce to factorial) it is suggestive that $\alpha\leq 1$ is the answer. I am having trouble upgrading this to the full $\Gamma$, as I don't really know much about this function.
My attempt:
EDIT: now I kinda solved it, but I'm looking for alternative method that are simpler and more accessible. My current solution is as follow:
Case $\alpha<0$: using Stirling's approximation (such as shown here: Quotient of gamma functions?) to show $\lim\limits_{x\rightarrow +\infty}\frac{\Gamma(x+\alpha)}{\Gamma(x)}=0$. Easily show that $\lim\limits_{x\rightarrow +\infty}x^{\alpha}=0$.
Case $\alpha=0$: direct calculation.
Case $0<\alpha<1$: apply Gautschi's inequality found here: How do you prove Gautschi's inequality for the gamma function?. After some substitution we can write it as  $(x+\alpha-1)^{\alpha}<\frac{\Gamma(x+\alpha)}{\Gamma(x)}<(x+\alpha)^{\alpha}$ with $0<\alpha<1$. Then show that $\lim\limits_{x\rightarrow\infty}((x+\alpha)^{\alpha}-(x+\alpha-1)^{\alpha})$ using mean value theorem: $(x+\alpha)^{\alpha}-(x+\alpha-1)^{\alpha}$ is the derivative of $x^{\alpha}$ at some point in $(x+\alpha-1,x+\alpha)$, which go to $0$. Then squeeze in.
Case $\alpha=1$: use the functional equation $\Gamma(x+1)=x\Gamma(x)$.
Case $\alpha>1$: the case $\alpha$ is integer is trivial: simply restrict domain to integer $x$ and rewrite as factorial. Otherwise first rewrite $\frac{\Gamma(x+\alpha)}{\Gamma(x)}=(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)\frac{\Gamma(x+\beta)}{\Gamma(x)}$ where $0<\beta=\alpha-[\alpha]<1$ (the bracket is floor). Then use Gautschi's inequality to get $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)\frac{\Gamma(x+\beta)}{\Gamma(x)}>(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)(x+\beta-1)^{\beta}$. Now we can show $\lim\limits_{x\rightarrow +\infty}x^{\alpha}-x^{[\alpha]}(x+\beta-1)^{\beta}=\lim\limits_{x\rightarrow +\infty}\frac{x^{\beta}-(x+\beta-1)^{\beta}}{x^{[\alpha]}}=0$ using L'Hospital. Hence the problem reduced to showing $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)(x+\beta-1)^{\beta}-x^{[\alpha]}(x+\beta-1)^{\beta}$ go to infinity. Since $(x+\beta-1)^{\beta}$ go to infinity, we just need to show $(x+\beta)(x+\beta+1)\ldots(x+\alpha-1)-x^{[\alpha]}$ stay away from $0$, which is simply just expanding out the term. This show that none of the $\alpha>1$ work.
However, this proof is awfully complicated, and relying on more obscure results. And it does not intuitively showed why the cut off point is at $\alpha=1$. I am looking for a better proof.
Thank you for your help.

A:

Let $\alpha \in \mathbb R$ be fixed.  We'll assume that $\alpha \neq 0$ and $\alpha \neq 1$ --- in these cases the behavior of $\frac{\Gamma(x+\alpha)}{\Gamma(x)} - x^\alpha$ is trivial anyway.

Stirling's formula for the gamma function tells us that
$$
\Gamma(x) = \sqrt{\frac{2\pi}{x}} \left(\frac{x}{e}\right)^x \left[1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)\right] \tag{1}
$$
as $x \to +\infty$.  If $x \to +\infty$ then $x+\alpha \to +\infty$, so we also have the formula
$$
\Gamma(x+\alpha) = \sqrt{\frac{2\pi}{x+\alpha}} \left(\frac{x+\alpha}{e}\right)^{x+\alpha} \left[1 + \frac{1}{12(x+\alpha)} + O\left(\frac{1}{(x+\alpha)^2}\right)\right], \tag{2}
$$
valid as $x \to +\infty$.  One can check that
$$
1 + \frac{1}{12(x+\alpha)} + O\left(\frac{1}{(x+\alpha)^2}\right) = 1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)
$$
as $x \to +\infty$.  This, together with a few algebraic manipulations of $(2)$, yields
$$
\Gamma(x+\alpha) = x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \sqrt{\frac{2\pi}{x}} \left(\frac{x}{e}\right)^x \left[1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)\right] \tag{3}
$$
as $x \to +\infty$.  By dividing $(3)$ by $(1)$ we obtain
$$
\begin{align}
\frac{\Gamma(x+\alpha)}{\Gamma(x)} &= x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \frac{1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)}{1 + \frac{1}{12x} + O\left(\frac{1}{x^2}\right)} \\
&= x^\alpha \left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} \frac{\left(1+\frac{\alpha}{x}\right)^x}{e^\alpha} \left[1 + O\left(\frac{1}{x^2}\right)\right] \tag{4}
\end{align}
$$
as $x \to +\infty$.

By the binomial theorem we have
$$
\left(1+\frac{\alpha}{x}\right)^{\alpha-1/2} = 1 + \frac{(\alpha-1/2)\alpha}{x} + O\left(\frac{1}{x^2}\right) \tag{5}
$$
as $x \to +\infty$.
To approximate the factor involving the exponential function we proceed like
$$
\begin{align}
\left(1+\frac{\alpha}{x}\right)^x &= \exp\left\{x \log\left(1+\frac{\alpha}{x}\right)\right\} \\
&= \exp\left\{x \left[\frac{\alpha}{x} - \frac{\alpha^2}{2x^2} + O\left(\frac{1}{x^3}\right)\right]\right\} \\
&= \exp\left\{\alpha - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right\} \\
&= e^\alpha \exp\left\{- \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right\} \\
&= e^\alpha \left[1 - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right], \tag{6}
\end{align}
$$
where the steps are valid as $x \to +\infty$.

Plugging $(5)$ and $(6)$ into $(4)$ yields
$$
\begin{align}
\frac{\Gamma(x+\alpha)}{\Gamma(x)} &= x^\alpha \left[1 + \frac{(\alpha-1/2)\alpha}{x} + O\left(\frac{1}{x^2}\right)\right] \left[1 - \frac{\alpha^2}{2x} + O\left(\frac{1}{x^2}\right)\right] \left[1 + O\left(\frac{1}{x^2}\right)\right] \\
&= x^\alpha \left[1 + \frac{\alpha^2-\alpha}{2x} + O\left(\frac{1}{x^2}\right)\right] \\
&= x^\alpha + \frac{\alpha^2-\alpha}{2} x^{\alpha-1} + O(x^{\alpha-2})
\end{align}
$$
as $x \to +\infty$.  Thus
$$
\frac{\Gamma(x+\alpha)}{\Gamma(x)} - x^\alpha \sim \frac{\alpha^2-\alpha}{2} x^{\alpha-1}
$$
as $x \to +\infty$, and the right-hand side tends to zero if and only if $\alpha - 1 < 0$.

