Relationship Between Differentiability and Continuity

Theorem. Every differentiable function is continuous.

Proof. Let $f(x)$ be differentiable at $x=a$, so

The claim to prove is that $f$ is continuous at $x=a$, i.e. $\displaystyle\lim_{x\to a}f(x)=f(a)$; writing $x=a+h$ (so $x\to a\iff h\to0$), this is the same as showing $\displaystyle\lim_{h\to0}f(a+h)=f(a)$.

Multiply both sides of (1) by $h$ (legitimate since $h\to0$ but $h\ne0$ throughout the limiting process):

so $\displaystyle\lim_{h\to0}f(a+h)=f(a)$, which is precisely the statement that $f$ is continuous at $x=a$. $\blacksquare$

The converse is false. A continuous function need not be differentiable — continuity is necessary but not sufficient for differentiability. This is shown by the standard example $f(x)=|x|$ on $R$, i.e. $f(x)=-x$ for $x<0$ and $f(x)=x$ for $x\ge0$.

Continuity at $x=0$: $\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x)=0$ and $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0$; also $f(0)=0$. Since both one-sided limits equal $f(0)$, $f$ is continuous at $x=0$.

Differentiability at $x=0$: the claim is that $f'(0)$ does not exist, i.e. $Lf'(0)\ne Rf'(0)$.

$Lf'(0)=\lim_{h\to0^-}\dfrac{f(0+h)-f(0)}h=\lim_{h\to0^-}\dfrac{-h-0}h=\lim_{h\to0^-}(-1)=-1$ …