Let f be differentiable on all of R. Prove that f is continuous on all of R.

Step-by-step solution

1. 1

   Reduce the statement to a single point. Continuity on $\mathbb{R}$ means continuity at every $a\in\mathbb{R}$, so it suffices to fix an arbitrary $a$ and show

   $\lim_{x\to a}f(x)=f(a)$

   Equivalently, $\displaystyle\lim_{x\to a}\big(f(x)-f(a)\big)=0$.

2. 2

   Use the hypothesis. By assumption $f$ is differentiable at $a$, which means the limit

   $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$

   exists as a finite number. That existence is the only thing available, so the proof must be arranged to use it.

3. 3

   Insert the difference quotient artificially. For $x\neq a$ the following is an identity, obtained by multiplying and dividing by $x-a$:

   $f(x)-f(a)=\frac{f(x)-f(a)}{x-a}\cdot(x-a)$

   This is the whole trick: it turns an unknown difference into a product of something whose limit is known and something that visibly tends to $0$.

4. 4

   Take limits with the product rule for limits. Both factors have limits, so the limit of the product is the product of the limits:

   $\lim_{x\to a}\big(f(x)-f(a)\big)=\left(\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\right)\cdot\left(\lim_{x\to a}(x-a)\right)=f'(a)\cdot 0=0$

   The finiteness of $f'(a)$ is essential — an infinite derivative would leave an indeterminate $\infty\cdot 0$.

5. 5

   Conclude. Therefore $\displaystyle\lim_{x\to a}f(x)=f(a)$, i.e. $f$ is continuous at $a$. Since $a$ was arbitrary,

   $\boxed{f\text{ is continuous on all of }\mathbb{R}}$

6. 6

   Note that the converse fails. Continuity does not imply differentiability: $f(x)=|x|$ is continuous at $0$ but the difference quotient $\tfrac{|x|}{x}$ tends to $+1$ from the right and $-1$ from the left, so $f'(0)$ does not exist. Weierstrass's function is continuous everywhere and differentiable nowhere, showing the gap is not a minor exception.