Differentiability and Continuity

The definitions above raise an obvious question: can a function be continuous at a point yet still fail to have a derivative there? The book answers this with a sequence of illustrations before proving the one implication that does always hold.

Illustration — a corner. $f(x)=|x-2|$ is continuous at $x=2$ (no break in the graph), but

Since $f'(2^-)=-1\ne1=f'(2^+)$, $f'(2)$ does not exist: the graph of $y=|x-2|$ has a sharp corner at $(2,0)$, with no single well-defined tangent line there. (At every other point $x_0\ne2$, $f'(x_0)=1$ if $x_0>2$ and $f'(x_0)=-1$ if $x_0<2$, so $f$ is differentiable everywhere except at the corner itself.)

Illustration — a vertical tangent. $f(x)=x^{1/3}$ is continuous everywhere (no hole or break), but at $x=0$,

which does not exist as a finite number, so $f$ is not differentiable at $x=0$ even though it is continuous there — the graph simply has a vertical tangent at the origin.

Example — the greatest integer function. $f(x)=\lfloor x\rfloor$ is not even continuous at any integer $n$, since $\lim_{x\to n^-}\lfloor x\rfloor = n-1 \ne n = \lim_{x\to n^+}\lfloor x\rfloor$; so $f'(n)$ cannot exist at any integer $n$ (differentiability needs continuity first, established below).

Illustration — a jump. For $f(x)=x$ if $x\le0$ and $f(x)=x+1$ if $x>0$, the graph jumps by $1$ at $x=0$ (a jump discontinuity), and correspondingly $f'(0^-)=1$ while $f'(0^+)$ blows up (the right-hand difference quotient $\to\infty$ as $\Delta x\to0^+$), so $f'(0)$ does not exist.

The three ways differentiability fails. These illustrations exhaust the possibilities: a function $f$ fails to be differentiable at $x_0$ (a point of its domain) exactly when one of the following holds — (i) $f$ has a vertical tangent at $x_0$; (ii) the graph comes to a point at $x_0$ (a sharp edge $\vee$ or peak $\wedge$ — a corner/cusp); or (iii) $f$ is discontinuous at $x_0$. In short: discontinuity always implies non-differentiability — but, as the $|x-2|$ and $x^{1/3}$ examples show, the converse direction (continuity implying differentiability) is emphatically false.

What does always hold is the one-directional implication in the other order:

Theorem 10.1 (Differentiability implies continuity). If $f$ is differentiable at $x=x_0$, then $f$ is continuous at $x_0$.

Proof. Since $f$ is differentiable at $x_0$, $f'(x_0)=\lim_{\Delta x\to0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}$ exists as a unique real number. Write the numerator as that same difference quotient multiplied back by $\Delta x$:

Taking the limit of both sides as $\Delta x\to0$ and using the product-of-limits rule, …