# Confirmation of proof that differentiability implies continuity?

I am trying to reason through the proof that differentiability implies continuity. I would appreciate some help reasoning through this one, if someone has a thorough explanation of the topic. I have tried to rewrite the definition of continuity to be equivalent to the definition of a derivative. My question is: does this work as a valid proof? If not, why? Thanks in advance!

Attempt at proof of concept.

• No, it doesn't; not every continuous function is differentiable. Dividing inside the limit by something that is going to zero may spoil the convergence entirely, or change the result to something else.
– Ian
Commented Apr 10, 2020 at 2:09
• A transparent way to do it is to say that $\lim_{x \to x_0} f(x)-f(x_0)=\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} (x-x_0)$ and now use the multiplication rule $\lim_{x \to a} f(x) g(x) = \left ( \lim_{x \to a} f(x) \right ) \left ( \lim_{x \to a} g(x) \right )$ when both limits on the right side exist.
– Ian
Commented Apr 10, 2020 at 2:10
• Thanks, @Ian. I think I may have been so preoccupied with wrangling it to use the "f(x+h)" form, I was missing the point. Thank you for the re-write. And yes, the 0/h side of the equation was why I asked the question originally, it was not satisfying. Could you please clarify your comment of "spoiling the convergency?" I do not quite understand what you mean by it. Commented Apr 10, 2020 at 13:20
• As an example, $\lim_{x \to 0} x \sin(1/x)=0$ but $\lim_{x \to 0} \frac{x \sin(1/x)}{x}$ doesn't exist. For another massive class of examples, consider just differentiation in general: by your logic, if a function is continuous at $a$ then it is differentiable at $a$ with derivative $0$, which is definitely not right! In these cases $h$ is going to zero at "just the right rate" so that $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists as a finite but nonzero number.
– Ian
Commented Apr 10, 2020 at 13:22
• Well put. Thanks @Ian! Commented Apr 10, 2020 at 15:30

As mentioned in the comments, if $$f$$ is diffentiable at $$x$$ then, $$\lim_{t \to x} \frac{f(t) - f(x)}{t-x} = f'(x)$$. So
$$\lim_{t\to x} f(t) - f(x) = \lim_{t \to x} \frac{f(t) - f(x)}{t-x}(t-x) = f'(x) \cdot 0$$
since both limits exist. Then given $$\epsilon > 0$$ from the definition of limit we can find a $$\delta > 0$$ such that $$t \in (x - \delta, x + \delta)$$ implies $$|f(t) - f(x)| < \epsilon$$. So $$f$$ is continuous at $$x$$.