Erroneous Proof that the Derivative is Continuous

continuityderivativesfake-proofsreal-analysis

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.

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 $$n\geq N$$.