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.

We show first that $f'(c)$ exists and is equal to $\lim_{x\to c}f'(x)$. Let $g$ be the function on $A$ given by $g(x)=f'(x)$ for $x\neq c$ and $g(c)=\lim_{x\to c}f'(x)$ which makes $g$ continuous. If $h$ and $k$ are small but positive then by the fundamental theorem of calculus we have $$f(c+h)-f(c+k)=\int_{c+k}^{c+h}g(t)dt.$$ Taking the limit for $k$ gives us $$f(c+h)-f(c)=\int_{c}^{c+h}g(t)dt.$$ Choosing appropriate $k$ we can get the same for negative $h$. It now follows that $$\left|\frac{f(c+h)-f(c)}{h}-g(c)\right|=\left|\frac{1}{h}\int_{c}^{c+h}g(t)dt-g(c)\right|=$$$$\left|\frac{1}{h}\int_{c}^{c+h}g(t)-g(c)dt\right|\leq \frac{1}{|h|}\int_{c}^{c+h}|g(t)-g(c)|dt.$$ Since $g$ is continuous we can choose $h$ small enough so that $|g(t)-g(c)|<\varepsilon$ in this interval. We now get $$\frac{1}{|h|}\int_{c}^{c+h}|g(t)-g(c)|dt\leq\frac{1}{|h|}|h|\varepsilon=\varepsilon$$ and we are done. Now we can apply this same logic to the function $f'(x)$ and so on.
