Calculate the limit of $\frac{n^{n+1}}{n!}\int_0^xe^{-nt}t^n\,\mathrm dt$ Set $R_n(x)$ the quantity in the title, for $n\in\mathbb{N}$ and $x > 0$. 
I'm trying to prove that if $x < 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} 0$, if $x > 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} 1$ and if $x = 1$, then $R_n(x)\underset{n\rightarrow +\infty}{\longrightarrow} \frac{1}{2}$. (I got these limits with Python.)
My approach is to make a substitution in the integral, $u = nt$. I get then $R_n(x) = \displaystyle\frac{1}{n!}\int_0^{nx}e^{-u}u^n\,\mathrm du$. I recognize the (incomplete) Gamma function, and know that $\Gamma(n+1) = n!$ but don't know how to proceed (I tried many other substitutions, without success).
Could I get some help on the matter? Thanks.
 A: A probabilistic proof: Apply the central limit theorem to independent exponentially distributed random variables $X_1,X_2,\dots\sim\operatorname{Exp}(\lambda=1)$.  Note that $X_1+\dots+X_n\sim \operatorname{Gamma}(\alpha=n, \beta=1)$ has mean $n$ and variance $n$.  So
$$
R_n(x)=\mathbb{P}(X_1+\dots+X_n\leq nx)\simeq\Phi((x-1)\sqrt{n})\to\begin{cases}
0 & x<1\\
1 & x>1\\
\frac12 & x=1
\end{cases}.
$$

If you don't like probability, you can directly appeal to asymptotic expansions such as
$$
\frac{\gamma(n+1,a+y(2n)^{1/2})}{\Gamma(n+1)}
=\frac12\operatorname{erfc}(-y)-\frac13\left(\frac2{n\pi}\right)^{1/2}(1+y^2)\exp(-y^2)+O(n^{-1})\quad (y\in\mathbb{R}, n\to\infty)
$$
found by Tricomi in 1950.
A: Here's a non-probabilistic casting of @user10354138's logic.
You want $$\lim_{n\to\infty}\frac{n^{n+1}}{n!}\int_0^x e^{-nt}t^ndt=\lim_{n\to\infty}\frac{1}{n!}\int_0^{nx}e^{-y}y^ndy.$$A bit of calculus gvies the quadratic approximation $n\ln y-y\approx n\ln n-n-\frac{(y-n)^2}{2n}$, so$$\int_0^{nx}e^{-y}y^ndy\approx n^ne^{-n}\int_0^{nx}e^{-(y-n)^2/(2n)}dy\approx\frac{n!}{\sqrt{2\pi n}}\int_{-n}^{n(x-1)}e^{-z^2/(2n)}dz,$$where the second $\approx$ uses the Stirling approximation. So we want$$\lim_{n\to\infty}\frac{1}{\sqrt{2\pi n}}\int_{-n}^{n(x-1)}e^{-z^2/(2n)}dz.$$If $x>1$, the integral $\approx\int_{-\infty}^\infty e^{-z^2/(2n)}dz=\sqrt{2\pi n}$, so the limit is $1$. If $x=1$, we lose the $z>0$ half of the integral, halving the result; if $x<1$, the upper limit $\to-\infty$, making the limit $0$.
