Calculate $\lim\limits_{n \to \infty}{nf^{(n)}}(x)$ where $xf(x)=e^x-1$ 
Given that $xf(x)=e^x-1$ calculate $\lim\limits_{n \to \infty}{nf^{(n)}}(x)$. 

I have try to figure the $n$th derivative of the function but it looks messy. The only useful recurrence I found is this one:
$$nf^{(n-1)}(x)+xf^{(n)}(x)=e^x$$
 A: Partial solution: feel free to edit to add the last missing piece (it's a Community Wiki answer).
I've had to refrain from using the power series for $\frac{e^x-1}{x}$, namely $\sum_{n=0}^\infty \frac{x^n}{(n+1)!}$, to answer. This is tempting, though, and I encourage you to try: in any case, this lets one show very easily that $$n f^{(n)}(0) = \frac{n}{n+1} \xrightarrow[n\to\infty]{} 1.$$
Start with your recurrence relation, specifically
$$
n f^{(n-1)}(x)+xf^{(n)}(x)=e^x\tag{for all $x$}
$$
which is easy to prove by induction. Switching indices, we get that for $n\geq 1$
$$
(n+1) f^{(n)}(x)+xf^{(n+1)}(x)=e^x\tag{for all $x$}
$$
and therefore
$$
n f^{(n)}(x)=\frac{n}{n+1}\left( e^x - xf^{(n+1)}(x) \right)\tag{for all $x$}
$$
If we knew that, for any fixed $x_0$, $\lim_{n\to\infty}f^{(n)}(x_0) = 0$, we'd then be done: indeed, then the RHS of the above relation would converge to $1\cdot (e^{x_0}-0)$, yielding that 
$$
\lim_{n\to\infty} n f^{(n)}(x) = e^x
$$
for any fixed $x$. We don't know that, however.
A: Hint:
$$f(x)=\int_0^1e^{xt}~\mathrm dt $$
Thus,
\begin{align}f^{(n)}(x)&=\int_0^1t^ne^{xt}~\mathrm dt\\&=\frac{e^x}{n+1}-x\int_0^1\frac{t^{n+1}}{n+1}e^{xt}~\mathrm dt\end{align}
where
$$0\le\left|\int_0^1\frac{t^{n+1}}{n+1}e^{xt}~\mathrm dt\right|\le\int_0^1\frac{t^{n+1}}{n+1}e^{|x|}~\mathrm dt=\frac{e^{|x|}}{(n+1)(n+2)}$$
