nth time differentiability at a point if limit of nth derivative exists at that point Exercise:
Let the function $f$ be defined and continuous in an open interval $A$. Suppose that $c$ is a point in $A$ and that $f$ has derivatives up to order $m$ on the set $A \backslash\{c\}$. Suppose further that $\lim\limits _{x \rightarrow c} f^{(k)}(x)$ exists for $k=1, \ldots, m$ and the limits are finite numbers. Show that $f$ has derivatives up to order $m$ in all of $A$. Moreover $f^{(k)}(c)=\lim\limits _{x \rightarrow c} f^{(k)}(x)$, for $k=1, \ldots, m$
I know proof of the case when $k=1$. It has a lot of answers in this site, for example here
for complete induction step I was suggested the following expression:
$$ f^{(k)}(c)=\lim _{x \rightarrow c} \frac{f^{(k-1)}(x)-f^{(k-1)}(c)}{x-c}=\lim _{x \rightarrow c} \frac{\int_{c}^{x} f^{(k)}(t) d t}{x-c}=\lim _{x \rightarrow c} f^{(k)}(x)$$
but we haven't studied integral yet. Is there any technique to replace integral with that one??
Thank you in advance.

As you have mentioned in your question, you are aware of the proof for the case when $k=1$. Let us state this version for $k=1$ explicitly.

Theorem: Let $A$ be an open interval and $f:A\to\mathbb {R} $ be continuous on $A$. Let $c\in A$ be such that $f$ is differentiable at all points of $A$ except possibly at $c$ and $\lim_{x\to c} f'(x) $ exists. Then $f'(c) $ exists and equals $\lim_{x\to c} f'(x) $.

Now let $f:A\to\mathbb {R} $ be continuous on $A$ and let $c\in A$ and $m$ be a positive integer such that $f^{k} (x) $ exists for all $x\in A\setminus \{c\} $ for $k=1,2,\dots,m$. And further assume that $\lim_{x\to c} f^{(k)} (x) $ exist for $k=1,2,\dots,m$.
Note that the above hypotheses imply that $f^{(k)}$ is continuous on $A\setminus \{c\} $ for all $k=1,2,\dots,m-1$. Using the theorem at the start of the answer we get that $f'(c) $ exists and equals the limit of $f'$ at $c$ so that $f'$ is continuous on $A$. Now apply the same theorem on $f'$ and conclude that $f''$ is continuous on $A$. Repeat the procedure a finite number of times to conclude that $f^{(m-1)}$ is continuous on $A$. And finally using the theorem once more we get that $f^{(m)}(c) $ exists and equals the limit of $f^{(m)} $ at $c$.
Observe that $f^{(m)} $ is continuous at $c$ but we can't say anything about continuity of $f^{(m)} $ at other points of $A$.
So you can see that the essentially the case $k=1$ is all we need.
