TITLE: Limit of derivative function
QUESTION [6 upvotes]: $f:\Bbb{R}\to\Bbb{R},$ $f$ continuous, defined by $xf(x) = e^x-1$ then $$\lim _{n\to \infty }nf^{\left(n\right)}\left(x\right)= ?$$
I've tried to calculate $f'(x), f''(x)$ and $f'''(x)$ but I didn't find any pattern.

REPLY [3 votes]: Though you may not be very familiar with Taylor's theorem, I intend to post this answer so that you may see how powerful it is.  Notice that
$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$
And thus,
$$f(x)=\sum_{k=0}^\infty\frac{x^k}{(k+1)(k!)}$$
$$f^{(n)}(x)=\sum_{k=0}^\infty\frac{x^k}{(k+1+n)(k!)}$$
$$nf^{(n)}(x)=\sum_{k=0}^\infty\frac{nx^k}{(k+1+n)(k!)}$$
And as $n\to\infty$, we end up with (quite nicely, if we take the limit through the sum)
$$\lim_{n\to\infty}nf^{(n)}(x)=\sum_{k=0}^\infty\frac{x^k}{k!}=e^x$$
A quick check says this should be right.

To take the limit through the sum, you should use uniform convergence, which follows through quite easily.