JOVANA
Library Glossary Getting Started Three Levels Fields How it works Mission
Join the mission
All guides

The epsilon–delta definition of a limit

What does “f(x) approaches L as x approaches a” really mean? We replace the vague picture with the precise epsilon–delta challenge, and prove one limit by hand.

From a picture to a promise

Informally, the limit of a function f at a point a is the value L that f(x) gets close to as x gets close to a. The trouble is the word “close.” Close enough for what? The genius of the epsilon–delta definition (function) is to turn this into a game with a guaranteed winning strategy. A skeptic names a tolerance epsilon > 0, demanding that f(x) land within epsilon of L. You must answer with a margin delta > 0 so that every x within delta of a (but not equal to a) is forced into that tolerance.

Formally: we say lim of f(x) as x -> a equals L when for every epsilon > 0 there exists delta > 0 such that for all x, if 0 < |x - a| < delta then |f(x) - L| < epsilon. Notice 0 < |x - a| excludes x = a itself: the limit never asks what happens at a, only near it. That is why a must be a cluster point of the domain — every neighborhood of a must contain other points of the domain, or the condition is vacuous.

Proving a limit by hand

The standard recipe is scratch work, then proof. In the scratch work you start from the target |f(x) - L| < epsilon and work backwards to find how small |x - a| must be. Then in the clean proof you announce delta first and verify forwards. Let us prove lim of (3x + 1) as x -> 2 equals 7.

Claim: lim_{x->2} (3x + 1) = 7.

Scratch work (find delta):
  |f(x) - L| = |(3x + 1) - 7| = |3x - 6| = 3|x - 2|.
  We want 3|x - 2| < epsilon, i.e. |x - 2| < epsilon/3.
  So delta = epsilon/3 should work.

Proof:
  Let epsilon > 0 be given. Choose delta = epsilon/3 > 0.
  Suppose 0 < |x - 2| < delta. Then
     |f(x) - 7| = |3x - 6| = 3|x - 2| < 3*delta = 3*(epsilon/3) = epsilon.
  Hence |f(x) - 7| < epsilon, as required.  QED
A linear limit: here delta = epsilon/3 works for every a, because the function has constant slope 3.

For a nonlinear function like x^2 the slope changes, so delta will depend on a as well as epsilon. A common trick is to first restrict delta <= 1 to bound the “bad” factor, then take delta = min(1, something/epsilon). The whole machinery rests on the absolute value as a distance and on its inequalities; mastering these manipulations is most of the battle in early real analysis.