# Epsilon-delta proof confusion

#### sweatingbear

##### Member
In a epsilon-delta proof, one wishes to show that $$\displaystyle 0 < |x-a| < \delta$$ implies $$\displaystyle |f(x) - \text{L}| < \epsilon$$. But every example that I have seen, one always begins with the "then"-part, namely the conseqeunt $$\displaystyle |f(x) - \text{L}| < \epsilon$$, and work backwards to arrive at something reminiscent of $$\displaystyle 0 < |x-a| < \delta$$.

But is this not the wrong way to carry out a proof of a conditional? From what I know, you conventionally have to begin with the antecedent (i.e. the "if"-part), assume it and see if it can lead you to the conclusion (and not the other way around). My spontaneous thought is that we do this because of the quantifiers "for every" and "there exists" but I am not sure.

Example: We wish to show that

$$\displaystyle \lim_{x \rightarrow 2} \, (x^2) = 4 \, .$$

We wish to show that if $$\displaystyle 0 < |x-2| < \delta$$ then $$\displaystyle |x^2 - 4| < \epsilon$$. Naturally, one would then start off with $$\displaystyle 0 < |x - 2| < \delta$$ and go on from there, but according to every example that I have seen that is not the case.

Can somebody help me see things clearer?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
In a epsilon-delta proof, one wishes to show that $$\displaystyle 0 < |x-a| < \delta$$ implies $$\displaystyle |f(x) - \text{L}| < \epsilon$$. But every example that I have seen, one always begins with the "then"-part, namely the conseqeunt $$\displaystyle |f(x) - \text{L}| < \epsilon$$, and work backwards to arrive at something reminiscent of $$\displaystyle 0 < |x-a| < \delta$$.

But is this not the wrong way to carry out a proof of a conditional? From what I know, you conventionally have to begin with the antecedent (i.e. the "if"-part), assume it and see if it can lead you to the conclusion (and not the other way around). My spontaneous thought is that we do this because of the quantifiers "for every" and "there exists" but I am not sure.

Example: We wish to show that

$$\displaystyle \lim_{x \rightarrow 2} \, (x^2) = 4 \, .$$

We wish to show that if $$\displaystyle 0 < |x-2| < \delta$$ then $$\displaystyle |x^2 - 4| < \epsilon$$. Naturally, one would then start off with $$\displaystyle 0 < |x - 2| < \delta$$ and go on from there, but according to every example that I have seen that is not the case.

Can somebody help me see things clearer?
Hi sweatingbear!

The key is that these proofs have to hold for any $\varepsilon > 0$.
So these proofs do start at the end.
Then you have to find a $\delta >0$ which is dependent on $\varepsilon$, for which the implication holds.

#### Ackbach

##### Indicium Physicus
Staff member
Hi sweatingbear!

The key is that these proofs have to hold for any $\varepsilon > 0$.
So these proofs do start at the end.
Then you have to find a $\delta >0$ which is dependent on $\varepsilon$, for which the implication holds.
And I would add that few people can see what $\delta$ needs to be from the beginning. So the typical work-flow is that, before you write the actual proof, you find your $\delta$ as a function of $\epsilon$. Then you write your actual proof, which starts by letting $\epsilon$ be greater than zero, then assumes $0<|x-a|< \delta$, and shows that $|f(x)-L|< \epsilon$.

The first bit where you find your $\delta$ is not actually part of the proof proper, and you could omit it. However, for pedagogical purposes, having this weird expression come out of nowhere would be confusing to students, and they would wonder how they could find it.

See here for a general way to find $\delta$ values, at least for values of a function that are not local extrema.

#### sweatingbear

##### Member
All right, that helped see things differently (hopefully clearer); thanks for your replies.

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