$f:\Bbb{R}\to\Bbb{R},$ $f$ 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.

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    $\begingroup$ How familiar are you with Taylor series? $\endgroup$
    – Arthur
    Jan 2, 2017 at 17:57
  • $\begingroup$ @Arthur Not at all... Haven't learned about them until now. If you could give me a hint using another method it would be appreciated. $\endgroup$
    – Liviu
    Jan 2, 2017 at 18:00
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    $\begingroup$ Is the fact that, for $n \geq 0$, $x f^{(n+1)} = e^x - (n+1) f^{(n)}$ of any help here? $\endgroup$
    – Dmoreno
    Jan 2, 2017 at 18:57
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    $\begingroup$ @Dmoreno It allows you to identify the limit, if you know it exists. If $nf^{(n)}$ converges to some finite value for each $x$ then, obviously, $f^{(n)}$ converges to $0$. So from the formula it follows that the limit is $e^x$, if it exists. $\endgroup$
    – Thomas
    Jan 2, 2017 at 19:46
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    $\begingroup$ @Liviu if you now that for fixed $x$, $nf^{(n)}(x)$ converges to some finite value $a$, then $f^{(n)}(x)$ behaves like $a/n$ which converges to $0$. So will $xf^{(n)}(x)$ (for the given $x$), and the limit of $nf^{(n)}(x)$ will coincide with the limit of $(n+1)f^{(n)}(x)$. Now use these results in the formula from Dmorenos comment. The missing link is the proof for the existence of the limit. $\endgroup$
    – Thomas
    Jan 2, 2017 at 20:14

1 Answer 1


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


And thus,




And as $n\to\infty$, we end up with (quite nicely, if we take the limit through the sum)


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.

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    $\begingroup$ In fact, it is easy to show that the limit is $e^x$ without appealing to uniform convergence or the dominated convergence theorem. $\endgroup$
    – Mark Viola
    Jan 2, 2017 at 19:28
  • $\begingroup$ @Dr.MV Actually, it was not as simple as I may have liked, so I think I'll leave it this way. $\endgroup$ Jan 2, 2017 at 19:43
  • $\begingroup$ Let $\epsilon >0$ be given. Choose $N$ large enough so that $\sum_{k=N}^\infty \frac{k+1}{n+k+1}\frac{x^k}{k!}<\epsilon/2$. Then, with that fixed $N$, take $n$ so large that the finite sum $\sum_{k=0}^ {N-1} \frac{k+1}{k+1+n}\frac{x^k}{k!}<\epsilon/2$. $\endgroup$
    – Mark Viola
    Jan 2, 2017 at 19:50

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