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 $a$.