How to prove asymptotic limit of an incomplete Gamma function How can I prove:
$$\lim_{z\to\infty}{\Gamma(z, x)\over\Gamma(z)} = 1$$
? Here $\Gamma(z, x)$ is the upper incomplete gamma function and $\Gamma(z)$ is the gamma function.
This must be something trivial, but I can't figure it out. Any ideas?
 A: The integrand $t^{z-1}\mathrm e^{-t}$ is unimodal and has its maximum at $t=z-1$. Thus for $z-1\gt x$ we can bound the integrand on both sides of $x$ by its value at $x$ to obtain
$$
\begin{align}
\Gamma(z)-\Gamma(z,x)
&=
\int_0^xt^{z-1}\mathrm e^{-t}\mathrm dt
\\
&\lt\int_0^xx^{z-1}\mathrm e^{-x}\mathrm dt
\\
&=xx^{z-1}\mathrm e^{-x}
\\
&=\frac x{z-1-x}\int_x^{z-1}x^{z-1}\mathrm e^{-x}\mathrm dt
\\
&\lt\frac x{z-1-x}\int_x^{z-1}t^{z-1}\mathrm e^{-t}\mathrm dt
\\
&\lt\frac x{z-1-x}\int_0^\infty t^{z-1}\mathrm e^{-t}\mathrm dt
\\
&=\frac x{z-1-x}\Gamma(z)\;,
\end{align}
$$
so for fixed $x$ the ratio of the difference to $\Gamma(z)$ tends to $0$ for $z\to\infty$.
A: In this answer I showed that
$$
\left(\frac{e^x}{e^x-1}\right)^z \gamma(z,x) \leq \Gamma(z)
$$
when $z \geq 1$ and $x > 0$, where
$$
\gamma(z,x) = \Gamma(z) - \Gamma(z,x)
$$
is the lower incomplete gamma function.  Rearranging we get
$$
0 \leq 1 - \frac{\Gamma(z,x)}{\Gamma(z)} = \frac{\gamma(z,x)}{\Gamma(z)} \leq \left(\frac{e^x-1}{e^x}\right)^z \longrightarrow 0
$$
as $z \to \infty$ when $x$ is fixed.
In fact, this shows that the result is still true if $x = f(z) \geq 0$ and
$$
z \log\left(1-e^{-f(z)}\right) \longrightarrow -\infty.
$$
In particular, it's still true if
$$
\frac{z}{e^{f(z)}} \longrightarrow \infty.
$$
