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I am wondering the following fact, and I believe I know the answer, but I am not sure why.

If $f(x)$ is a smooth function from $\mathbb{R}$ to $\mathbb{R}$, if $f(0)=0$, is it true that $f(x)/x$ is smooth?

I believe this is an application of Taylor's theorem (a text I was reading used Taylor's theorem at some part of a proof, and I believe it to be this part), however, I do not see why.

Any help is much appreciated.

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up vote 2 down vote accepted

Taylor's theorem with integral remainder at order 0 (this is a very pendantic way of calling the fundamental theorem of calculus) yields $$ f(x) = f(0) + \int_0^x f'(t)\,dt = f(0) + x\int_0^1f'(ux)\,du $$

Hence using $f(0)=0$, $$ \frac{f(x)}{x} = \int_0^1 f'(ux)du $$ is smooth.

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Thanks much. That is really easy. :-) – user45150 Jan 24 '13 at 7:34
Why is the RHS smooth? – user7530 Jan 24 '13 at 7:38
It is smooth because $f^{(n)}$ is continuous as $f$ is smooth so we may bring any order of derivative through the integral. – user45150 Jan 24 '13 at 7:40

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