Differentiable Implies Continuous

Differentiable implies continuous, but not the other way around. This scene zooms into the corner that shows why.

Continuous only means the graph has no breaks. Differentiable asks for more. The tangent slope must settle to a single value. At a sharp corner the slope splits into two.

For example, a 🚗 car climbing a smooth hill has a speed at every moment. But on a path with a sharp zigzag corner, the speed at the corner is not one single value.

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}
lim⁡h→0f(a+h)=f(a)\lim_{h \to 0} f(a+h) = f(a)

Part of these learning paths

Examples & Exercises

Prove in three lines: if ff is differentiable at aa, it is continuous at aa.

f(a+h)−f(a)=f(a+h)−f(a)h⋅h→f′(a)⋅0=0f(a+h) - f(a) = \frac{f(a+h)-f(a)}{h} \cdot h \to f'(a) \cdot 0 = 0
The full ∣x∣|x| — the proof's trick is multiplying the difference quotient by hh.
Show proof

  1. For h≠0h \neq 0, f(a+h)−f(a)=f(a+h)−f(a)h⋅hf(a+h) - f(a) = \frac{f(a+h)-f(a)}{h} \cdot h — dividing by a nonzero number and multiplying back changes nothing.

  2. As h→0h \to 0, the bracketed difference quotient tends to f′(a)f'(a) and the trailing hh tends to 00, so the product tends to f′(a)⋅0=0f'(a) \cdot 0 = 0.

  3. Hence lim⁡h→0f(a+h)=f(a)\lim_{h \to 0} f(a+h) = f(a) — the definition of continuity at aa, so ff is continuous at aa.

Prove f(x)=∣x∣f(x) = \lvert x\rvert is not differentiable at x=0x = 0.

∣h∣h={1h>0−1h<0\frac{|h|}{h} = \begin{cases} 1 & h > 0 \\ -1 & h < 0 \end{cases}
Zoom level 4 — left slope −1-1, right slope +1+1, unchanged.
Show proof

  1. The difference quotient is f(0+h)−f(0)h=∣h∣h\frac{f(0+h)-f(0)}{h} = \frac{|h|}{h}, which equals 11 for h>0h > 0 and −1-1 for h<0h < 0.

  2. So the right-hand limit is 11 and the left-hand limit is −1-1; they differ, hence the two-sided limit — the derivative — does not exist.

  3. That is why the corner never flattens under zoom: at every magnification the left slope stays −1-1 and the right slope stays +1+1.

State (as a theorem) that a function exists which is continuous everywhere but differentiable nowhere.

Idea level; proof omitted.

Zoom level 8 — not one corner of ∣x∣|x|, but such a corner at every point.
Show hint

  1. Weierstrass's theorem: there exists a function continuous at every point yet differentiable at none. A classic example is a series of the form ∑n=0∞ancos⁡(bnπx)\sum_{n=0}^{\infty} a^n \cos(b^n \pi x) with 0<a<10 < a < 1, bb an odd positive integer, and ab>1+3π2ab > 1 + \frac{3\pi}{2}.

  2. The idea: a uniform limit of continuous functions is continuous, which secures continuity everywhere, but at each point the jagged waves defeat every difference quotient, so no derivative exists anywhere. No proof is given — this is stated as a known result.

10. Differentiation