Differentiability and Continuity

By the end of this lesson, you will be able to state and apply the theorem relating differentiability and continuity, identify non-differentiable points, and understand why differentiability is stronger than continuity.

The Key Theorem

If $f$ is differentiable at $c$, then $f$ is continuous at $c$.

Contrapositive: if $f$ is NOT continuous at $c$, then $f$ is NOT differentiable at $c$.

The converse is FALSE: $f$ can be continuous but NOT differentiable.

Why the Converse Fails

$f(x) = |x|$ at $x = 0$:

Continuous: $f(0) = 0 = \lim_{x\to 0}|x|$ ✓

Differentiable? $\lim_{h\to 0^+}h/h = 1$ and $\lim_{h\to 0^-}(-h)/h = -1$

Left ≠ right → not differentiable.

Types of Non-Differentiable Points

Corner: the function is continuous but the left and right derivatives are unequal. (|x| at 0, |x-2| at 2)

Cusp: the left and right derivatives both approach $\pm\infty$ but disagree in sign. ($f(x) = x^{2/3}$ at 0)

Vertical tangent: $f'(x) \to \pm\infty$ (both sides go to same infinity). ($f(x) = x^{1/3}$ at 0)

Discontinuity: any discontinuity destroys differentiability.

Example 1 — Checking Differentiability at a Piecewise Boundary

$$f(x) = \begin{cases}x^2 & x < 1 \ 2x-1 & x \geq 1\end{cases}$$

Continuity at $x = 1$: $\lim_{x\to 1^-} = 1$ and $\lim_{x\to 1^+} = 1 = f(1)$ ✓

Differentiability:

Left: $f'(x) = 2x \to 2$ as $x \to 1^-$

Right: $f'(x) = 2$ for $x \geq 1$

Left derivative = right derivative = 2 → differentiable at $x = 1$ ✓

Example 2 — Non-Differentiable at a Corner

$$g(x) = \begin{cases}x^2 & x \leq 0 \ 2x & x > 0\end{cases}$$

Continuous at $x = 0$: $g(0) = 0 = \lim_{x\to 0} g(x)$ ✓

Left derivative: $2x|_{x=0} = 0$; Right derivative: $2$; $0 \neq 2$ → not differentiable at $x = 0$.

Proof: Differentiability Implies Continuity

If $f'(c) = \lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ exists, then:

$\lim_{h\to 0}[f(c+h)-f(c)] = \lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}\cdot h = f'(c)\cdot 0 = 0$