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.
 A: The mean value theorem will allow us to escape a proof by integration. Assume the result holds for $k-1.$
Suppose in our open interval we have the point $b,$ $c<b,$ where $c$ is the point you mention. For $c<x<b,$ consider
$$\tag 1 \frac{f^{(k-1)}(x)-f^{(k-1)}(c)}{x-c}.$$
Now $f^{(k-1)}$ is continuous on $[c,b]$ and differentiable on $(c,b).$ By the MVT, $(1)$ equals $f^{(k)}(c_x).$ As $x\to c^+,$ $c_x\to c^+.$ Therefore, by hypothesis, $f^{(k)}(c_x)$ converges to a limit I'll call $L_k.$ We thus have shown the right hand derivative of $f^{(k-1)}$ at $c$ is $L_k.$
The same idea works for $b<c.$ Thus the left hand derivative of $f^{(k-1)}$ at $c$ is also $L_k.$ We have shown $[f^{(k-1)}]'(c)= f^{(k)}(c)=L_k.$ And since we know $\lim_{x\to c}f^{(k)}(x)=L_k=f^{(k)}(c),$ we see that $f^{(k)}$ is continuous at $c.$
So this is how a proof by induction will work to give the full result.
A: for the induction step, apply the base case to $f^{(m)}$
example from m=1 to m=2
f'(c) exists and f' is cts at x=c.
$f^{(2)}$ exists on $A\setminus\{c\} \implies$ f' is diff on $A\setminus\{c\} \implies$ f' is cts on $A\setminus\{c\}$
f' is cts on A.
apply base case to f'
$f^{(2)}(c)$ exists and $f^{(2)}$ is cts at x=c.
A: 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.
A: 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.
