I have written the following which argues that for any arbitrary sequence of points such that $x_n \to c, x_n\neq c$ we must also have $f'(x_n)\to f'(c)$ which would imply that $f'(x)\to f'(c)$ as $x\to c$ , i.e., the derivative is continuous. However, I can't see where the error in my argument is.

Let $f:[a,b] \to \mathbb{R}$ be a differentiable function and $c\in [a,b]$.

Fix $\epsilon > 0$. By assumption, for $c\in [a,b]$, $f'(c)$ exists so put $\delta_1 > 0$ so small that whenever $0<\lvert x -c \rvert < \delta_1$ we must have

$$ \left \lvert \frac{f(x)-f(c)}{x-c}-f'(c) \right \rvert < \frac{\epsilon}{2} $$

$[a,b]$ is closed so there exists a sequence such that $x_n \to c$ and $x_n \neq c$. By assumption, $f'(x_n)$ exists so there exists a $\delta_2 >0$ such that whenever $0<\lvert x-x_n \rvert < \delta_2$ we must have

$$ \left \lvert f'(x_n)-\frac{f(x)-f(x_n)}{x-x_n} \right \rvert < \frac{\epsilon}{2} $$

Now pick an integer $N$ so large that for $n\geq N$ we must have $0<\lvert x_n – c\rvert < \mathrm{min}(\delta_1,\delta_2)$.

It follows that

$$ \left \lvert \frac{f(x_n)-f(c)}{x_n-c}-f'(c)\right \rvert < \frac{\epsilon}{2} \quad \textbf{and} \quad \left \lvert f'(x_n)-\frac{f(x_n)-f(c)}{x_n-c}\right \rvert < \frac{\epsilon}{2} $$

Applying the triangle inequality, we see that $\lvert f'(x_n)-f'(c) \rvert < \epsilon$ and we are done.

## Best Answer

Let me write it up so the question is not unanswered...

Note that your $\delta_2$ depends on $x_n$. So we should really write $\delta_2(n)$. That is for a fixed $n$, $\delta_2(n)$ is a positive real such that if $0\lt |x-x_n|\lt \delta_2(n)$, then $\left|\frac{f(x)-f(x_n)}{x-x_n} - f'(x_n)\right| \lt \epsilon/2$.

Then $\delta=\min(\delta_1,\delta_2(n))$ guarantees that $|f'(c)-f'(x_n)|\lt \epsilon$ for that

fixed$n$, not for an arbitrary $n$.So in order to ensure that this inequality holds for all $n\geq N$, you would need to find a $\delta$ that satisfies $\delta\leq \inf(\{\delta_1\}\cup\{\delta_2(n)\mid n\geq N\})$. Unfortunately, that infimum could be equal to $0$, which means that you have no guarantee that you can pick a $\delta\gt 0$ that ensures both inequalities you have at the end for

all$n\geq N$.