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This is a try to fill the holes in this previous exercise.

Let $f$ defined as $f(x):=(x-a)^n g(x)$, for some $g$ continuous in a neighborhood of $a$, we can say that $f$ is $n$ times differentiable at $a$?

If we define the quotient difference operator

$$\Delta^n_h f(x):=\frac1{h^n}\sum_{k=0}^n\binom{n}{k}f(x+(n-k)h)(-1)^k$$

then if $f$ is continuous in a neighborhood of $a$ and if Im not wrong we can write

$$\lim_{h\to 0}\Delta^n_h f(a)=f^{(n)}(a)$$

For the case of the title this reduces to

$$\begin{align}f^{(n)}(a)&=\lim_{h\to 0}\frac1{h^n}\sum_{k=0}^n\binom{n}{k}(-1)^k((n-k)h)^ng(a+(n-k)h)\\&=\lim_{h\to 0}\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a+(n-k)h)\\&=\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a)\end{align}$$

and for derivatives for $j\le n$ we have that $f^{(j)}(a)=0$.

It is this reasoning correct?

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    Not quite - your limit formula is true if we know that the limit in question exists, but it doesn't guarantee that the derivative exists a priori. – 2017-01-14
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    @πr8 thank you, this is exactly the kind of answer that I was searching for. – 2017-01-14
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    Typographical note: The AMS `align*` environment can be used to typeset wide display math so the "$=$"'s line up. If the backquote construction permits preformatted text, for example: `\begin{align*} f(x) &= g(x) \\ &= h(x). \end{align*}` (It doesn't, but perhaps the resulting hash is legible enough to be helpful anyway....) – 2017-01-14

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The reasoning is not correct. The equality $$\lim\limits_{h\rightarrow 0}\Delta_h^nf(a)=f^{(n)}(a)$$ is true, provided that the RHS exists. However, just because the limit on the LHS exist, it doesn't follow the RHS exists. Indeed, the statement in the question is false. This is because $f(x)$ can oscillate very rapidly in the neighbourhood of $a$, which would make the first derivative be discontinuous at $a$. To be precise, consider for example $$g(x)=x\sin x^{-2}\text{ for }x\neq 0,\\g(0)=0$$ (check it's continuous) and $$f(x)=x^2g(x)=x^3\sin x^{-2}.$$

We then find $$f'(x)=3x^2\sin x^{-2}-2\cos x^{-2},\\ f'(0)=0,$$ which is discontinuous at zero, hence $f$ is not twice differentiable.

For general $n$, you can similarly take $f(x)=x^{n+1}\sin x^{-n}$.

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    Or let $g$ be nowhere differentiable, then $(x-a)^n g(x)$ is (for $n \geqslant 1$) differentiable at $a$, but nowhere else. – 2017-01-14
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    Yup, this looks better than my thing :-) – 2017-01-14
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    @DanielFischer That works as well, and is probably quite easier to think of. – 2017-01-14