Validity of a statement on a function being infinitely differentiable at a point Thinking about a problem, i had got this idea which appears true, but i feel it is a direct consequence of a well known theorem, the name of which i would like to ask here.
The idea can be put in the form of a theorem as below.
If a function $f : \mathbb{R} \to \mathbb{R}$ is infinitely differentiable at a point $x = x_o$ then given any arbitrarily large $r \in \mathbb{N}$ we can find an $\epsilon \in \mathbb{R}$ such that $f$ is $\mathcal{C}^r$ in the interval $(x_o-\epsilon,x_o+\epsilon)$.
I'd like to know the validity of this statement and if it is valid, from which theorem does it follow ?
 A: The original post seems to be generating an excessively long string of comments. I will try to explain things using the more generous editing facilities of the Answers format.
Take any function $g(x)$.  Let us think about what we mean when we write
$$\lim_{x \to a}\ g(x)=b$$
Very crudely, we mean that whenever $x$ is close enough to $a$ (but not necessarily equal to $a$), $g(x)$ is close to $b$.  In order for the phrase  "$g(x)$ is  close to $b$" to have meaning, $g(x)$ must exist, that is, $g(x)$ must be defined at $x$, it must make sense at $x$.  So we conclude that if $\lim_{x\to a} \ g(x)=b$, then in particular $g(x)$ must make sense, must be a number, for all $x$ in some interval about $a$, except possibly at $x=a$.
Now let us look at derivatives.  In general, denote the $n$-th derivative of $f$ by $f^{(n)}$ (the $0$-th derivative of $f$ is defined to be $f$).
The $(k+1)$-th derivative of $f$ at $a$ is defined as follows.
$$f^{(k+1)}(a)=\lim_{x\to a}\frac{f^{(k)}(x)-f^{(k)}(a)}{x-a}$$
Look at this definition.  It mentions $f^{(k)}(a)$, so in particular, in order for $f^{(k+1)}(a)$ to exist, the very meaning of $f^{(k+1)}(a)$ forces $f^{(k)}(a)$ to exist. Also, the limit could not exist unless $f^{(k)}(x)$ existed for all $x$ close enough to $a$ but not equal to $a$.  So if $f^{(k+1)}(a)$ exists, then $f^{(k)}(x)$ must exist for all $x$ close enough to $a$, including $x=a$.  
To sum up, analysis of the very definition (meaning) of differentiation shows that if $f$ is $k+1$ times differentiable at $a$, then $f$ must be $k$ times differentiable in some neighbourhood of $a$. 
