Proving $\lim_{x\to\infty}\frac{\Gamma(x)}{\Gamma(x+\varepsilon)}=0$ for all $\varepsilon>0$. I want to show that, for all $\varepsilon>0$, we have $$\lim_{x\to\infty}\frac{\Gamma(x)}{\Gamma(x+\varepsilon)}=0.$$ This is a real-valued special case of 8.328, 2. in the eighth edition of Gradshteyn-Ryzhik.
For my attempts, see my answers below. In one of my answers, I also show that, as $x\to\infty$, $$\bbox[5px,border:2px solid #C0A000]{\frac{\Gamma(x+1)x^\varepsilon}{\Gamma(x+1+\varepsilon)}=1+O(1/x),}$$ which is a stronger statement that the one written above.
 A: Proposition. Let $f:]0,\infty[\to\mathbb ]0,\infty[$ be a function that satisfies

*

*$\ln\circ f$ is convex;

*$\lim_{x\to\infty}\frac{\ln\circ f(x)}{x}=\infty$.

Then $\lim_{x\to\infty}\frac{f(x)}{f(x+\varepsilon)}=0$ for every $\varepsilon>0$.

Note that 1. is true for the Gamma function (cf. Proofwiki) and 2. is true for the Gamma function by this.

Proof of Proposition. By log convexity, for any $\lambda\in[0,1]$ and $a,b\in]0,\infty[$, $$\ln\circ f(\lambda a+(1-\lambda) b)\le \lambda\ln(f(a))+(1-\lambda)\ln(f(b)).$$
So $$f(\lambda a+(1-\lambda) b)\le f(a)^\lambda f(b)^{1-\lambda}.$$ Set $a=1, b=x+\varepsilon, \lambda=\frac{\varepsilon}{\varepsilon+x-1}$, then you get $$f(x)\le f(1)^{\text{something}\le 1} f(x+\varepsilon)^{1-\varepsilon/(\varepsilon+x-1)}.$$
Therefore, for $C=\sup_{y\in[0,1]}f(1)^y\in\mathbb R$, we get $$\frac{f(x)}{f(x+\varepsilon)}\le C f(x+\varepsilon)^{-\varepsilon/(\varepsilon+x-1)}.$$
The logarithm of the right-hand side is (the inequality being true whenever $f(x+\varepsilon)\ge 1$, which is the case for $x$ large enough by the second condition on $f$) $$-\frac{\varepsilon}{\varepsilon+x-1}\ln(f(x+\varepsilon))\le -\varepsilon\frac{f(x+\varepsilon)}{x+\varepsilon},$$
which converges to $-\infty$ as $x\to\infty$ by the second condition on $f$, and then taking exponentials, one can conclude that the original term must go to $0$.
A: This is the same idea as in your answer, only with a slightly shorter proof.
$g(x) = \ln(\Gamma(x))$ is convex with $\lim_{x \to \infty} g(x)/x = \infty$. The convexity condition for $1 < x < x+\epsilon$ gives
$$
 g(x) \le \frac{\epsilon}{x+\epsilon-1}g(1) + \frac{x-1}{x+\epsilon-1}g(x+\epsilon)
$$
or
$$
g(x)-g(x+\epsilon) \le \frac{\epsilon}{x+\epsilon-1} \left( g(1) - g(x+\epsilon) \right) \, .
$$
It follows that $\lim_{x \to \infty} g(x)-g(x+\epsilon) = -\infty$, which is equivalent to $\lim_{x\to\infty}\frac{\Gamma(x)}{\Gamma(x+\epsilon)}=0.$.
A: One can also show the stronger (this is the real-valued Gradshteyn-Ryzhik statement)
Fact. For any fixed $\varepsilon>0$, $$\lim_{x\to\infty}\frac{\Gamma(x) x^\varepsilon}{\Gamma(x+\varepsilon)} = 1.$$ In fact, we even have, as $x\to\infty$, $$\bbox[5px,border:2px solid #C0A000]{\frac{\Gamma(x+1)x^\varepsilon}{\Gamma(x+1+\varepsilon)}=1+O(1/x).}$$
For the proof of this fact I will use the following Stirling-bound:
Theorem ([1; Theorem 1.6]). There exist real numbers $0<a<b$ such that, for all $x\ge 1$, $$\left(\frac xe\right)^x\sqrt{2\pi(x+a)}<\Gamma(x+1)<\left(\frac xe\right)^x\sqrt{2\pi(x+b)}.$$
Remark. Furthermore, it is proven in [1; Theorem 1.6] that the before-mentioned inequality holds for $a=\frac 16$ and $b=\frac{e^2}{2\pi}-1$, and that these constants cannot be improved.
It should be noted that one may be able to use more elementary Stirling approximations for $\Gamma$ in order to prove the result here. The one above (which is not easy to prove) is chosen only for convenience.
Proof of Proposition. $\varepsilon>0$ shall be fixed. For $x\ge 1$, let $$f(x)=\Gamma(x+1)\left(\frac xe\right)^{-x} \frac1{\sqrt{2\pi}\sqrt{x}}.$$
By the Theorem before, we have $f(x)\in\left]\sqrt{1+\frac ax},\sqrt{1+\frac bx}\right[$ for all $x$. In particular, $\lim_{x\to\infty} f(x)=1$ (in fact, $f(x)=1+O(1/x)$ as $x\to\infty$).
So we get $$\frac{\Gamma(x+1)x^\varepsilon}{\Gamma(x+1+\varepsilon)} = \color{blue}{\frac{\sqrt{x}}{\sqrt{x+\varepsilon}}\frac{f(x)}{f(x+\varepsilon)}} e^\varepsilon \color{green}{\frac{x^{x+\varepsilon}}{(x+\varepsilon)^{x+\varepsilon}}}.$$
We note first that, from the before argument for $f$, and an argument for the square root (for instance the Laurent series or by noting $g(x)\overset{\text{Def.}}=\sqrt{1+\varepsilon/x}-1=O(1/x)$ and then writing $\frac{1}{\sqrt{1+\varepsilon/x}}=\frac{1}{1+g(x)}=\frac{x}{x+xg(x)}=1-\frac{xg(x)}{x+xg(x)}$ and then noticing that $xg(x)=O(1)$)
$$\color{blue}{\frac{\sqrt{x}}{\sqrt{x+\varepsilon}}\frac{f(x)}{f(x+\varepsilon)}}=1+O(1/x).$$
The green term is equal to $$\color{purple}{\frac{x^x}{(x+\varepsilon)^x}}\color{orange}{\frac{x^\varepsilon}{(x+\varepsilon)^\varepsilon}}.$$
The purple term equals (for the convergence speed see this) $$\left(1+\frac\varepsilon x\right)^{-x}=e^{-\varepsilon}+O(1/x).$$
The orange term is $\left(1+\frac\varepsilon x\right)^\varepsilon$, which has the first order Laurent series $1+O(1/x)$. In conclusion, $$\bbox[5px,border:2px solid #C0A000]{\frac{\Gamma(x+1)x^\varepsilon}{\Gamma(x+1+\varepsilon)}=1+O(1/x)}$$ as $x\to\infty$, which is what we wanted to prove.
Literature
[1] Necdet Batir: Inequalities for the Gamma function. Arch. Math. 91, pages 554–563 (2008). https://doi.org/10.1007/s00013-008-2856-9
A: For the question in title
$$\lim_{x\to\infty}\frac{\Gamma(x)}{\Gamma(x+\varepsilon)}$$ take logarithms, use Stirling approximation and continue with Taylor series to obtain
$$\log\Bigg[\frac{\Gamma(x)}{\Gamma(x+\varepsilon)}\Bigg]=-\epsilon  \log (x)-\frac{(\epsilon -1) \epsilon }{2 x}+\frac{(\epsilon -1)
   \epsilon  (2 \epsilon -1)}{12 x^2}+O\left(\frac{1}{x^3}\right)$$ that is to say
$$\frac{\Gamma(x)}{\Gamma(x+\varepsilon)}=x^{-\epsilon }\Bigg[1-\frac{(\epsilon -1) \epsilon }{2 x}+\frac{(\epsilon -1) \epsilon  (\epsilon +1)
   (3 \epsilon -2)}{24 x^2}+O\left(\frac{1}{x^3}\right)\Bigg]$$
Doing the same for
$$\frac{\Gamma(x+1)\,x^\varepsilon}{\Gamma(x+1+\varepsilon)}=1-\frac{\epsilon  (\epsilon +1)}{2 x}+\frac{\epsilon  (\epsilon +1) (\epsilon +2)
   (3 \epsilon +1)}{24 x^2}+O\left(\frac{1}{x^3}\right)$$
A: From the identity $$\Gamma(x+1) = x\Gamma(x)\tag{$*$}$$ we have
$$\frac{\Gamma(x)}{\Gamma(x+\varepsilon)} = \frac{1}{x}\color{blue}{\frac{\Gamma(x+1)}{\Gamma(x+\varepsilon)}} \tag{1}$$
Applying Gautschi's inequality to the $\color{blue}{\text{blue}}$ expression, we observe that
$$\bbox[5px,border:2px solid #C0A000]{x^{-\varepsilon} < \frac{\Gamma(x)}{\Gamma(x+\varepsilon)} < (x+1)^{-\varepsilon}\left(1 + \frac{1}{x}\right)}$$
Now, just the Sandwich theorem will finish the job as $x\to \infty$ and $\varepsilon > 0$. $\quad \square$
Note: Gautschi's inequality gives the statement used for $\varepsilon \in (0, 1)$ only. However, the identity $(*)$ reduces everything back to $\varepsilon \in (0,1)$ (i.e. no loss of generality).
