# Have I found a counterexample in this question?

If $f:X\rightarrow \mathbb{R} \,$ is a function with $x_0 \in \overline{X} \,\setminus \partial(\overline{X})$ such that :

$$\exists \,\,\,\,f'_-(x_0)=\lim_{x\rightarrow x_0^-}\dfrac{f(x)-f(x_0)}{x-x_0},$$

$$\exists \,\,\,\,f'_+(x_0)=\lim_{x\rightarrow x_0^+}\dfrac{f(x)-f(x_0)}{x-x_0}$$ but with possibly $f'_-(x_0) \not= f'_+(x_0)$, does this still imply continuity of $f$ ?

if so then why can I have a function such as $$f(x) = \begin{cases} 3x \, , \text{ if} \,\,\, x<x_0 \\ 10x+1 \, ,\text{ if} \,\,\, x=x_0 \\ -2x \, , \text{ if} \,\,\, x>x_0 \end{cases}$$ that does have $f'_-(x_0)=3$ and $f'_+(x_0)=-2$ but is not continuous, does that mean that only the existence of the left and right side derivatives on a point do not guarantee that $f$ is continuous at $x_0$ ?

You haven’t found the limits right. I will take the example of $x_0=0$. There $$f’^+=\lim_{x\rightarrow 0^+} \frac{2x-1}{x} =\lim_{x\rightarrow 0^+} 2 - \frac{1}{x} = -\infty$$ and $$f’^-=\lim_{x\rightarrow 0^+} \frac{3x-1}{x} =\lim_{x\rightarrow 0^+} 3 - \frac{1}{x} = -\infty$$
Actually, continuity is implied because $\lim_{x\rightarrow x_0^+} f(x) = \lim_{x\rightarrow x_0^-}f(x) = f(x_0)$ since the two one-sided limits of your question exist (and are finite).
• I think you cannot use the value of the function at point $x_0$ to calculate the limit in general. Dec 6, 2017 at 18:17
Since the limits exist, you have $$\lim_{x\to x_0^+}f(x)= \lim_{x\to x_0^+}\left(\frac{f(x)-f(x_0)}{x-x_0}(x-x_0)+f(x_0)\right)=f(x_0)$$ and similarly for the limit from the left.
You are confusing $$f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$$ with $$\lim_{x\to x_0^+}f'(x)$$ which can be different.