About a limit with Euler $\Gamma$ function. For all $x \in \Bbb R_+^*$, we put:  $$f(x)=\frac{1}{\Gamma(x)}\int_x^{+\infty}t^{x-1}e^{-t}dt.$$
Can we compute the limit : $\displaystyle\lim_{x \to +\infty} f(x) $?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\fermi\pars{x}\equiv{1 \over \Gamma\pars{x}}
     \int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t:\ {\large ?}}$


First, we have to derive an asymptotic behavior of the integral when
  $\ds{x \gg 1}$. The derivation is similar to the one which leads to the Gamma function asymptotic behavior ( namely, the Stirling approximation ):

\begin{align}&\left.\dsc{\int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t}\,
\right\vert_{\,x\ \gg\ 1}
=\int_{x}^{\infty}\exp\pars{\bracks{x - 1}\ln\pars{t} - t}\,\dd t
\\[5mm]&\sim\int_{x}^{\infty}
\exp\pars{\bracks{x - 1}\ln\pars{x - 1} - \bracks{x - 1} - {\bracks{t - x + 1}^{2} \over 2\bracks{x - 1}}}\,\dd t
\\[5mm]&=\pars{x - 1}^{x - 1}\expo{-\pars{x - 1}}
\int_{1}^{\infty}\exp\pars{-t^{2} \over 2\bracks{x - 1}}\,\dd t
\\[5mm]&=\bracks{\pars{x - 1}^{x - 1}\expo{-\pars{x - 1}}}\root{2\pars{x - 1}}
\int_{1/\root{2\pars{x - 1}}}^{\infty}\exp\pars{-t^{2}}\,\dd t
\\[5mm]&\sim\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}\root{2}\
\overbrace{\int_{0}^{\infty}\exp\pars{-t^{2}}\,\dd t}^{\dsc{\root{\pi} \over 2}}
\\[5mm]&=\dsc{\half}\,
\bracks{\root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}}
\end{align}

Then

\begin{align}&\color{#66f}{\large%
\left.{\int_{x}^{\infty}t^{x - 1}\expo{-t}\,\dd t \over \Gamma\pars{x}}\,
\right\vert_{\,x\ \gg\ 1}}
\sim{\pars{\dsc{1/2}}\bracks{%
\root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}}\over
\root{2\pi}\pars{x - 1}^{x - 1/2}\expo{-\pars{x - 1}}}
\color{#66f}{\large\to\ \half}
\end{align}

The $\ds{\Gamma\pars{s,z}}$ asymptotic behavior as given in
  this link is misleading because $\ds{\tt\mbox{it should be valid for fixed}}$ $\ds{s}$. In another words,
  $\ds{\lim_{x\ \to\ \infty}\lim_{s\ \to\ x}\Gamma\pars{s,x}}$ can not be evaluated with such expansion which was the one I used blindly in my previous answer.

Thanks to @Mohamed who call my attention to this point and to
this paper and thanks to @anorton who was worried about the whole procedure. Thanks to both of them.
A: Does this help?
$\int_{0}^{\infty}t^{x-1}e^{-t}dt=\int_{0}^{x}t^{x-1}e^{-t}dt+\int_{x}^{\infty}t^{x-1}e^{-t}dt$
so 
$\int_{x}^{\infty}t^{x-1}e^{-t}dt=\Gamma\left(x\right)-\int_{0}^{x}t^{x-1}e^{-t}dt$
In taking the limit we get zero:
$\lim_{x\rightarrow\infty}\frac{\Gamma\left(x\right)-\int_{0}^{x}t^{x-1}e^{-t}dt}{\Gamma\left(x\right)}=1-1=0$
