# Existance of $n$-th derivative at $x_0$ and continuity at $x_0$

While reading my own scribbles from calculus I've noticed that when proving Taylor series we've assumed that if $f: \mathbb R \to \mathbb R$ which has $n$-th derivative that $f$ is continuous in a neighborhood of $x_0$ and not only at $x_0$. If that's true it would be really nice to hear a good explanation as to why it has to be this way.

I could not quite get why, because while thinking about the function $f(x)= x^3D(x)$ where $D(x)$ is the Dirichlet function I though that it might have a second derivative as well.

But if the above mentioned $f(x)$ is continuous not only at $x_0$ but in some area of $x_0$ then I could theoretically say that looking at $g(x)= x^{-3}$ which is continuous at all $\mathbb R \setminus 0$ that by limits arethmetics $g(x)*f(x)$ is continuous at $x_0 \neq 0$ which means that $D(x)$ is continuous somewhere and it's obviously not true.

Well the first derivative exists because $\lim \limits_{h\to 0} \frac{ f(h)-f(0)}{h} =0$ the second I'm pretty sure that exists as well if not I'd really like to understand why.

I've tried looking at limits of this but could not quite get something concrete.

So to better summarize my questions, is it true that :

1. $f: \mathbb R \to \mathbb R$ which has $n$-th derivative $\Rightarrow$ $f$ is continuous in a neighborhood of $x_0$.
2. $f(x)= x^nD(x)$ where $D(x)$ is the Dirichlet function and $n \geq2 \in \mathbb N$ $\Rightarrow$ $f(x)$ has $n-1$-th derivative.

P.S this is not actual homework but it's related to the material so I've tagged it as homework just in case.

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For the Taylor series, you use the Mean value theorem, this one is only true if the function is differentiable in the intervall $(x_0,x_0+h)$.
Your expamle is differentiable, but only in $x=0$, so you can't say that $$f(x+h)=f(x)+f'(\xi) h$$ with $\xi\in (x,h)$

The normal definition of differentiable is that a function $f:I\rightarrow \mathbb{R}$ is differentiable in the point $x$ of the interior of $I$ when the limit of $$\frac{f(x+h)-f(x)}{h}$$ This one only makes sense when $f(x+h)$ exists. As $f$ is differentiable in $x_0$ implies $$f(x_0+h)=f(x_0)+h f'(x_0) +r(h)$$ where $$\lim_{h\rightarrow 0} \frac{r(h)}{h} =0$$ We can see over sequences that $f$ is continuous in $x_0$. As the second derivative only can exists when $f'(x+h)$ exists we know that when a function is 2 times differentialbe must be continuous in a neighbourhood of $x_0$.

As the Derivative of $x^n D(x)$ only exists in $1$ point, it is not 2 times differentiable.

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And while a derivative on an interval need not be continuous in general, existence of $f^{(n+1)}$ on an interval implies continuity of $f^{(n)}$ on that interval. – Jonas Meyer Feb 20 '13 at 15:29
Did I missunterstood the question? – Dominic Michaelis Feb 20 '13 at 15:33
I understood that my questions were not too clear so I've edited the question hope it's better. Sorry. – Scis Feb 20 '13 at 15:38
is it better now ? – Dominic Michaelis Feb 20 '13 at 15:50
Thanks, Yes it is, just two small things : 1). I think you forgot to complete your sentence at "when the limit of" 2). Could you explain in a little more detail why $f'(x+h)$ has to exist not only at $x_0$ in order to be two times differentiable, thanks again. – Scis Feb 20 '13 at 16:29