MT2002 Analysis

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## The epsilon-delta definition

From the above definition of convergence using sequences is useful because the arithmetic properties of sequences gives an easy way of proving the corresponding arithmetic properties of continuous functions. We now use this definition to deduce the more well-known ε-δ definition of continuity.

Informally: Close points (δ apart) are mapped to close points (ε apart).

We formalise this to get the following:

Definition

A function f from R to R is continuous at a point p ∈ R if
given ε > 0 there exists δ > 0 such that if |p - x| < δ then |f (p) - f (x)| < ε.

Theorem
This definition is equivalent to the previous one.

Proof
First, we show that continuity in this definition implies continuity in the previous definition.
So suppose (xn)→ p. We must prove that (f (xn))→ f (p).
That is given ε > 0 we must find N such that |f (xn) - f (p)| < ε if n > N.
Given such an ε from the new definition of continuity there is a δ such that |f (xn) - f (p)| < ε whenever |xn- p| < δ.
Since (xn) converges to p, there is a N such that this happens whenever n > N and so with this value of N our definition is satisfied.

Secondly, we show that the sequential definition implies the ε-δ one.
Given ε > 0, suppose that we could not find a suitable δ. Then δ = 1 would not work and so we must have some x1 such that |x1- p| < 1 and |f (x1) - f (p)| > ε.
Similarly, δ = 1/2will not work, and so we can find x2 further down the sequence than x1 such that |x2- p| < 1/2 and |f (x2) - f (p)| > ε.
Continuing in this way we get a sequence (x1, x2, x3, ... ) which by construction converges to p, but for which f (xn) is always at least ε away from f (p). So we cannot have (f (xn)) converging to f (p) and we have a contradiction. Application

The function defined by f (x) = √x is continuous.

Proof
Given ε > 0 we must show that |√x - √p| < ε provided that x, p are close enough.
Now |√x - √p| = |x - p|/|√x + √p| < |x - p| /√p and so choosing δ = ε/√p will do. Remark

The definition of continuity "doesn't quite work" at p = 0 for this function since the function is not defined for x < 0. One can define a notion of "one-sided continuity" to take care of examples like this.

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JOC September 2001