Tropic of Calculus » Lesson 1.7: Proofs with Delta-Epsilon

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Lesson 1.7: Proofs with Delta-Epsilon

We have just acquainted ourselves with the general concept of delta-epsilon, or the use of error bounds to define a limit. Now we set about the goal of proving a particular limit by providing a function delta(epsilon) — “you give me an epsilon, and I’ll give you a delta”.

Recall that our specific goal is to prove the following:

$\forall\varepsilon>0\:\exists\delta>0:0 is less than |x-a| is less than \delta\Rightarrow|f(x)-L| is less than \varepsilon$

In plain English, for every positive epsilon there must exist a positive delta such that being within delta of implies that f(x) is within epsilon of ; then and only then can we say that lim(x->a)f(x)=L .

We will return to our previously-examined function f(x)=2x+1 . Our goal is to prove that lim(x->2)f(x)=5 . First, we must guess a value for delta . We begin with epsilon representing a given positive number, and we wish to find a delta such that |(2x+1)-5| is less than epsilon whenever 0 is less than |x-2| is less than delta .

We rewrite |(2x+1)-5| as follows: |(2x+1)-5|=|2x-4|=2|x-2| , so our problem becomes finding delta such that 2|x-2| is less than epsilon whenever 0 is less than |x-2| is less than delta . Equivalently, we want delta such that |x-2| is less than (1/2)epsilon whenever . This suggests that we should guess delta=(1/2)epsilon .

Now that we have accomplished this, we must prove that our delta works. We restate our given:

Given

, we choose delta=(1/2)epsilon

and therefore delta>0

If 0 is less than |x-2| is less than delta then we work with the expression for epsilon , |(2x+1)-5|=2|x-2| is less than epsilon . Since we know that epsilon=2*delta , we rewrite the last expression as 2|x-2| is less than 2*delta , or . We have already shown that delta>0 from our given, so we have shown that delta=(1/2)epsilon implies 0 is less than |x-2| is less than delta . Therefore, by the definition of a limit, we have proven that lim(x->2)f(x)=5 .