Continuity and the ε–δ Definition

30 min
0/4 practice checks

A function ff is continuous at aa if we can force f(x)f(x) as close to f(a)f(a) as we like by keeping xx close enough to aa. Precisely, the ε\varepsilonδ\delta definition:

ε>0    δ>0    such that    xa<δ    f(x)f(a)<ε.\forall\, \varepsilon > 0 \;\; \exists\, \delta > 0 \;\; \text{such that} \;\; |x - a| < \delta \;\Rightarrow\; |f(x) - f(a)| < \varepsilon.

Again a challenge–response game: the adversary sets an output tolerance ε\varepsilon; you must supply an input window δ\delta that keeps the output inside ε\varepsilon.

Worked example. Show f(x)=3x+1f(x) = 3x + 1 is continuous at every aa.

Given ε>0\varepsilon > 0, notice f(x)f(a)=3x3a=3xa|f(x) - f(a)| = |3x - 3a| = 3\,|x - a|. Choose δ=ε3\delta = \dfrac{\varepsilon}{3}. Then xa<δ|x-a| < \delta gives f(x)f(a)=3xa<3δ=ε|f(x)-f(a)| = 3|x-a| < 3\delta = \varepsilon. Done — one clean choice of δ\delta works for every aa.