$\lim_{x \to \infty} \Gamma(x+1)/\Gamma(x+1+1/x^2)$ Can someone please help me with the calculation of this limit?
$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{\Gamma(x+1+1/x^2)}$$
I tried wolframalpha and seems to be 1, yet there are no "detailed steps" as to how it reaches the conclusion (I understand there wouldn't be even if I pay, as in other cases it shows me the first ones).
Thanks in advance,
Sergio
 A: $$\Gamma(z+\epsilon) =\Gamma(z) + \epsilon \Gamma'(z) + O(\epsilon^2) $$
$$\Gamma'(z) = \Gamma(z) \psi(z) $$
so the limit is
$$\lim_{x \to \infty} \frac1{1+ \psi(x+1)/x^2} = \lim_{x \to \infty} \frac1{1+ (\log{x}-\gamma)/x^2} = 1$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to \infty}\,\,{\Gamma\pars{x + 1} \over \Gamma\pars{x + 1 + 1/x^{2}}}} =
\lim_{x \to \infty}\,\,{x! \over \pars{x + 1/x^{2}}!}
\\[5mm] = &\
\lim_{x \to \infty}\,\,{\root{2\pi}x^{x + 1/2}\,\,\expo{-x} \over \root{2\pi}\pars{x + 1/x^{2}}^{x\ +\ 1/x^{2}\ + 1/2}
\,\,\,\,\expo{-x - 1/x^{2}}\,\,}
\\[5mm] = &\
\lim_{x \to \infty}\,\,{x^{x + 1/2} \over
x^{x\ +\ 1/x^{2}\ + 1/2}\,\,\,
\bracks{\pars{1 + 1/x^{3}}^{x^{3}}}^{1/x^{2}\ +\ 1/x^{5}\ + 1/\pars{2x^{3}}}
\,\,\,\,}\,\expo{1/x^{2}}
\\[5mm] = &
\lim_{x \to \infty}x^{-1/x^{2}} =
\exp\pars{-\lim_{x \to \infty}{\ln\pars{x} \over x^{2}}} =
\exp\pars{-\lim_{x \to \infty}{1/x \over 2x}}
\\[5mm] = &
\bbx{1} \\ &
\end{align}
