I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$

Differentiability Definition

When we say a function is differentiable at $x_0$, we mean that the limit:

$$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.

Continuity Definition

When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$

Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.


Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} = ‎f^{\prime} ‎(x) $$

and hence

$$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$

We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.

My Question

The proof seems to execute the following steps:

  1. Assume the function is continuous at $x_0$
  2. Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.

What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.

Hopefully my confusion is clear. Any help will be greatly appreciated.


Technically, there is an implicit issue of existence of limits which is being swept under the rug in the presentation you have given. The assumption of differentiability at $x_0$ says that the limit

$$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$

exists as a finite number. The limit $\lim_{x \to x_0} x-x_0$ exists and is zero regardless of our assumptions. Then the product rule for limits tells us both that $\lim_{x \to x_0} f(x)-f(x_0)$ exists, and that it is the product of the two limits above, which means it must be zero. Because the product rule also tells us that the limit exists, we do not have to assume continuity first.

  • 3
    $\begingroup$ I like your way. So we start with $\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0} = a$, where $a$ is finite. Using the product rule for limits, we show that $\lim_{x \to x_0} (f(x) - f(x_0)) = a \cdot \lim_{x \to x_0} (x - x_0) = 0$. Is that a correct interpretation? $\endgroup$
    – user2321
    Jun 6 '15 at 15:56
  • $\begingroup$ @user2321 That's right. $\endgroup$
    – Ian
    Jun 6 '15 at 15:59
  • $\begingroup$ Yes thank you, that is the kind of proof I was expecting. I will accept this answer. You have been very helpful, thanks! $\endgroup$
    – user2321
    Jun 6 '15 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.