The Epsilon-Delta Definition of a Limit
State what a limit claims with no appeal to motion: for every tolerance around the limiting value there is a distance around the point that keeps outputs inside it. Prove a simple limit by producing that distance.
What a learner can do afterwards
- Read an epsilon-delta statement as a challenge and a response
- Produce a delta that works for a given epsilon on a linear function
- Explain why the definition has to hold for every epsilon, not one small one
1 · Read
You claim that f(x) tends to L as x tends to a. Your challenger picks any positive tolerance epsilon around L. You must answer with a positive distance delta around a so that every x within delta of a, except a itself, lands with f(x) within epsilon of L.
Take f(x) = 3x + 1 near x = 2, where the limit is 7. The output error is |3x + 1 - 7| = 3|x - 2|. For any epsilon, delta = epsilon / 3 works, since 3|x - 2| then stays below epsilon.
One small epsilon is never enough, since the challenger may demand any closeness at all. The definition says for every epsilon, so your answer must cover each demand. Note that 0 < |x - a| leaves out x = a, since limits track nearness, not the value at the point.
For a line f(x) = mx + b, divide by the slope: delta = epsilon / |m|. Check by putting it back: |f(x) - L| becomes |m| times |x - a|, which lands below epsilon. For y = x squared near 0, x squared < epsilon means |x| < sqrt(epsilon), so delta = sqrt(epsilon).
Epsilon is the demanded closeness, delta is your answer, and the pair must work for every epsilon.
2 · Watch
Take it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.