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$f:\Bbb{R}\to\Bbb{R},$ $f$ 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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    How familiar are you with Taylor series? – 2017-01-02
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    @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. – 2017-01-02
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    May be [Leibniz rule for derivative of a product](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwiW-brwhqTRAhVW5mMKHeenCrkQFggaMAA&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FGeneral_Leibniz_rule&usg=AFQjCNHIN7Zv_Uz5hRAwCPGlsMel3dZj8w&bvm=bv.142059868,d.cGc) – 2017-01-02
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    @Nil Isn't there any other form of this learned in high school ? – 2017-01-02
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    Is the fact that, for $n \geq 0$, $x f^{(n+1)} = e^x - (n+1) f^{(n)}$ of any help here? – 2017-01-02
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    @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. – 2017-01-02
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    @Thomas Could you explain this once again ? And why isn't the limit $e^x/x$ if $f^{\left(n\right)}$converges to 0 – 2017-01-02
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    @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. – 2017-01-02
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    @Thomas Thank you very much, this really helped me ! – 2017-01-02

1 Answers 1

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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.

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    In fact, it is easy to show that the limit is $e^x$ without appealing to uniform convergence or the dominated convergence theorem. – 2017-01-02
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    @Dr.MV Actually, it was not as simple as I may have liked, so I think I'll leave it this way. – 2017-01-02
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    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 $\sum_{k=0}^ {N-1} \frac{k+1}{k+1+n}\frac{x^k}{k!}<\epsilon/2$. – 2017-01-02