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The fundumental theorem of calculus states:
if :
1) $f$ was integrable over an interval like $[a,b]$.
2) $f$ was continuous at $x=c$, $a<c<b$.
3) $F(x)= \int_a ^x{f(t)}$.

then:
$F'(c)=f(c)$

I wonder if there is any function with following properties :

1) $f$ was integrable over an interval like $[a,b]$.
2) $f$ was continuous over $[a,b]$ but at $x=c$, where isn't continuous.
3) $F(x)=\int_a^x{f(t)}$, and $F$ is differentiable over $[a,b]$, and $F'(x)=f(x)$

Note:

if there was any function like $f$ with above properties, then, $L_1$ and/or $L_2$ wouldn't exist ((event in the cases $L_1=\pm\infty$ and/or $L_2=\pm\infty$)), where:
$$L_1=\lim_{x \rightarrow c^-} f(x) \quad L_2=\lim_{x \rightarrow c^+}f(x)$$
any response, would be appreciated.

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I think the following function covers the properties you need.

$f(x) = \begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}), x \neq 0 \\ 0, x=0 \end{cases}$

It is continuous in $[a,b]$ where $a$ is a negative solution of the equation $2x\sin(\frac{1}{x})-\cos(\frac{1}{x})=0$ and $b$ a positive number, except from $x=c=0$ (that's where it is not continuous).

It is integrable over $[a,b]$.

$F(x)=\int_a^x f(t)dt =\begin{cases} x^2\sin(\frac{1}{x}), x \neq 0 \\ 0, x=0 \end{cases} $

and $F'(x)=f(x)$ for every $x \in [a,b]$.

As you have noted, $L_1$ and $L_2$ do not exist.

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