Calculus 13: Continuity versus Differentiability

There is a close relationship between continuity and differentiability, namely

Theorem. If $f'(x_0)$ exists then $f(x)$ is continuous at $x_0$; i.e. $\displaystyle\lim_{x\to x_0}f(x)=f(x_0)$. However the converse need not be true.

Proof. \begin{align*}\lim_{x\to x_0}[f(x)-f(x_0)]&=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)\\&=f'(x)\cdot 0\\&=0.\end{align*}

Example. [A Counterexample for the Converse] The function $f(x)=|x|$ is continuous at $x=0$ but has no derivative at $x=0$.

Proof. \begin{align*}\lim_{x\to 0+}\frac{f(x)-f(0)}{x-0}&=\lim_{x\to 0+}\frac{|x|}{x}\\&=\lim_{x\to 0+}\frac{x}{x}\\&=1,\end{align*} while \begin{align*}\lim_{x\to 0-}\frac{f(x)-f(0)}{x-0}&=\lim_{x\to 0-}\frac{|x|}{x}\\&=\lim_{x\to 0-}\frac{-x}{x}\\&=-1.\end{align*} Hence, $f'(0)=\displaystyle\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.

The graph of $y=|x|$