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Epsilon-Delta Proof

A proof of a formula on limits based on the epsilon-delta definition. An example is the following proof that every linear function f(x)=ax+b (a,b in R,a!=0) is continuous at every point x_0. The claim to be shown is that for every epsilon>0 there is a delta>0 such that whenever |x-x_0|<delta, then |f(x)-f(x_0)|<epsilon. Now, since

|f(x)-f(x_0)|=|ax+b-(ax_0+b)|]
(1)
=|ax-ax_0|
(2)
=|a|·|x-x_0|,
(3)

it is clear that

|x-x_0|<epsilon/(|a|) implies |f(x)-f(x_0)|<|a|·epsilon/(|a|)=epsilon.
(4)

Hence, for all epsilon>0, delta=epsilon/|a|>0 is the number fulfilling the claim.

See also

Epsilon, Epsilon-Delta Definition

This entry contributed by Margherita Barile

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Cite this as:

Barile, Margherita. "Epsilon-Delta Proof." From MathWorld--A Wolfram Resource, created by Eric W. Weisstein. https://mathworld.wolfram.com/Epsilon-DeltaProof.html

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