Here's the well known proof:

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

a) Therefore, $\lim_{x \to c} f(x) = f(c)$, $f$ is continuous at $c$.

The problem
I know how the equation works, but I don't get how it proves explicitly that differentiability implies continuity. To me, as long as $\lim_{x \to c} (x - c) = 0$, then why should $f ^{\prime} (x)$ even matter? a) would be remain true anyways, right?

I must be looking at it wrong. To me, it just looks like an equation saying that continuity implies continuity. As long as $\lim_{x \to c} (x - c) = 0$, which is the continuity equation, then of course the equation would imply that f(c) is continuous. But how does differentiability explicitly imply continuity?

It just looks like modifications were made to the continuity equation to make the differentiability equation appear, then it just went back to the continuity equation. But how exactly does the differentiability equation that popped up imply continuity?

I mean, it just looks like the continuity equation proving itself, but I don't see how the differentiability equation is "proving" anything. It started from the continuity equation and it went back to the continuity equation?

I'm so frustrated with this. I keep attending the same classes and the teachers seem to see something I can't.

  • 3
    $\begingroup$ What if $\frac{f(x)-f(c)}{x-c}$ goes to $\infty$ as $x$ goes to $c$? Namely, if $f$ is not differentiable at $x = c$, there's no guarantee that this doesn't happen. $\endgroup$
    – Michael L.
    Jul 27 '17 at 3:32
  • $\begingroup$ If that happens, wouldn't that mean that the slope becomes infinity? Wouldn't that be an equation like x = 10? What would be the equation for f(x)? $\endgroup$
    – user467501
    Jul 27 '17 at 3:38
  • 1
    $\begingroup$ If the slope becomes $\infty$ then $f$ is not differentiable at $x = c$ as @MichaelLee said above. This is because $0 \times \infty$ is undefined. Also, no. if $f(x) = 10$, then $f'(x) = 0$. $\endgroup$
    – Skam
    Jul 27 '17 at 3:42
  • $\begingroup$ Ah I see how the proof is useful now. Are there any other cases where the equation above will show that it isn't differentiable, or only infinity cases? $\endgroup$
    – user467501
    Jul 27 '17 at 3:44
  • $\begingroup$ The strategy really only fails when the derivative is unbounded near the point in question. Otherwise, the limit above can be bounded by a constant times the limit of $x-c$. $\endgroup$
    – Michael L.
    Jul 27 '17 at 3:55

Consider the function $$ \varphi(x)= \begin{cases} \dfrac{f(x)-f(c)}{x-c} & x\ne c \\[6px] \,f'(c) & x=c \end{cases} $$ which is defined in a neighborhood of $c$ and continuous at $c$, because $f$ is differentiable at $c$ by assumption.

You surely agree that also the function $$ \psi(x)=(x-c)\varphi(x)+f(c) $$ is continuous at $c$, because it's obtained from $\varphi$ by multiplication with a continuous function and addition of a constant. Now, by definition, $$ \psi(x)= \begin{cases} (x-c)\dfrac{f(x)-f(c)}{x-c}+f(c) & x\ne c \\[6px] (x-c)\cdot f'(c)+f(c) & x=c \end{cases} $$ and therefore $\psi(x)=f(x)$. This proves $f$ is continuous at $c$.

The above is just a different way of looking at the proof you already have. The key, in both arguments, is the existence of the derivative at $c$, from which we can deduce continuity.

This doesn't mean a function needs to be differentiable in order for being continuous. There is a famous example by Weierstrass of a function defined over $\mathbb{R}$ which is everywhere continuous but nowhere differentiable.

The proof you're given with uses a theorem about limits.

Theorem. Suppose $p$ and $q$ are defined in a punctured neighborhood of $c$ and that $$ \lim_{x\to c}p(x)=P \qquad\textit{and}\qquad \lim_{x\to c}q(x)=Q\tag{*} $$ (with $P,Q\in\mathbb{R}$). Then $$ \lim_{x\to c}p(x)q(x)=PQ $$

The theorem does not allow to argue anything about the limit of $p(x)q(x)$ if one of the two limits in (*) does not exist (or is infinite). It is used in an essential way in the proof you have, namely in $$ \lim_{x \to c} (x - c) \frac{f(x) - f(c)}{x - c}= \left[\lim_{x \to c} (x - c)\right]\left[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c}\right] $$ You may want to look at the argument backwards: \begin{align} 0&=0\cdot f'(c) \tag{1}\\[6px] &=\left[\lim_{x \to c} (x - c)\right]\left[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c}\right] \tag{2}\\[6px] &=\lim_{x \to c} (x - c) \frac{f(x) - f(c)}{x - c} &&\text{by the theorem} \tag{3}\\[6px] &=\lim_{x\to c}\bigl(f(x)-f(c)\bigr) &&\text{by algebra}\tag{4} \end{align} Without differentiability, you're not allowed to do the step from $(2)$ to $(3)$.

In other words, differentiability is a sufficient condition for continuity. But a function can as well be continuous at a point without being differentiable at the point, as the example of $$ f(x)=\begin{cases} x\sin\dfrac{1}{x} & x\ne0 \\[6px] 0 & x=0 \end{cases} $$ shows. If you try to work out the steps $(1)$–$(4)$ above with this function (at $c=0$), you'll see that you can't perform the step from $(2)$ to $(3)$.

  • $\begingroup$ For the second equation, $ψ(x)$, why is $(x-c) \cdot 0 + f(c)$ when x = c, instead of $0 \cdot f^{\prime}(c) + f(c)$ when x = c? When $x = c$, would the $(x-c) $ become 0? The rest of your answer really helped me understand the proof. $\endgroup$
    – user467501
    Jul 27 '17 at 23:30
  • $\begingroup$ @imagentab Because of a mistake! Thanks for noting, now it's fixed. $\endgroup$
    – egreg
    Jul 27 '17 at 23:44

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