How to Prove a Limit Using Delta/Epsilon Proofs

[Nidhogg]

I'm looking for a good, general explanation of how to do delta/epsilon proofs. I've searched all over the web but this stuff is just really confusing. Let me start with a problem, show my attempt at a solution, and then maybe you guys can explain it in a way that will make sense to me.

Homework Statement

Prove that: $$\lim_{x \rightarrow 2} 3x + 1 = 7$$

Homework Equations

If $$(0 < |x-a| < \delta)$$ implies $$(|f(x) - L| < \epsilon)$$, then $$\lim_{x\rightarrow a} f(x) = L$$.

The Attempt at a Solution

We want to prove that: $$\lim_{x \rightarrow 2} 3x + 1 = 7$$ To do this, we must show that $$(0 < |x- 2| < \delta)$$ implies: $$(|3x + 1 - 7| < \epsilon)$$ So I start by taking: $$3x + 1 = 7$$ and simplifying it to $$3x - 6 = 0$$ Which we can set to being less than epsilon as a way of choosing a delta, thus $$|3x - 6| < \epsilon$$ which means that $$3|x-2| < \epsilon$$ so I can choose $$\delta = \frac{\epsilon}{3}$$. Now I can assume that $$0 < |x - 2| < \delta$$, and since $$\delta = \frac{\epsilon}{3}$$ this means that $$0 < |x - 2| < \frac{\epsilon}{3}$$

At this point, I just draw a blank. I feel as if I have everything I need to complete the proof, but I'm missing something, and all the explanations I read don't seem to help. Can someone please help me get this through my thick skull?

 

[ciubba]

Now I can assume that

0>|x−2|>δ​

zero is greater than delta? Doesn't that contrast with

If

(0<|x−a|<δ)​

?

This is essentially saying that you have a negative distance between x and a, which is confounding.

This is a good website for a better instructional:

http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

 

[RUber]

The goal is to show that for any fixed number epsilon, you can choose a delta such that $| x- 2| < \delta \implies |f(x) - 7| < \epsilon$.
You have done a great job in the algebra, but you lost some of the theory.
You have $| x- 2 | < \delta$ and $3| x-2 | < \epsilon$.
When you choose $\delta = \epsilon/3$, you are solving the equality not the inequality.
By putting the inequality back in, you should have shown that for any fixed epsilon, there exists a delta that satisfies the relationship.

 

[Nidhogg]

ciubba: D'oh, that was a typo! I fixed it.
RUber: Thanks, man. I'll work on it a little more tonight.

 

[Nidhogg]

So according to what RUber said, I have $|x - 2| < \delta$ and $3|x - 2| < \epsilon$. I need to find from these two that such a delta exists for every epsilon, and I do this by putting together an inequality that relates delta and epsilon?

So, to take another whack at it I have: $$3\delta = \epsilon$$ and I have $$|x - 2| < \delta$$ so can't I derive $$3|x - 2| < \epsilon$$ which means that $|3x - 6| < \epsilon$ which means that $|f(x) - L| < \epsilon$? In that case, haven't I completed the proof?

 

[ciubba]

So according to what RUber said, I have $|x - 2| < \delta$ and $3|x - 2| < \epsilon$. I need to find from these two that such a delta exists for every epsilon, and I do this by putting together an inequality that relates delta and epsilon?

So, to take another whack at it I have: $$3\delta = \epsilon$$ and I have $$|x - 2| < \delta$$ so can't I derive $$3|x - 2| < \epsilon$$ which means that $|3x - 6| < \epsilon$ which means that $|f(x) - L| < \epsilon$? In that case, haven't I completed the proof?

You just proved that when δ=ϵ/3, $$|3x +1-7| < \epsilon$$ In other words, that value of delta will only get you back to the "epsilon" inequality IF it is valid. The delta heavy proofs tend to have the form of find a value of delta, then prove that that delta is true, which you have just done. These proofs tend to be more useful for proving general cases than specific ones.

 

[Nidhogg]

Ciubba: I'm not sure, but I think I'm getting closer to grasping this. We know now that when δ=ϵ/3, the limit inequality of |3x + 1 - 7| < ϵ holds. So what else do I have to do? Do I have to prove that there exists a δ such that δ=ϵ/3 for all ϵ? Isn't that necessarily true anyway since ϵ is a real number implicitly? What do you mean by an epsilon value's being "valid?" What does it mean to prove that a delta is true?

Sorry I'm so dense.

 

[ciubba]

Ciubba: I'm not sure, but I think I'm getting closer to grasping this. We know now that when δ=ϵ/3, the limit inequality of |3x + 1 - 7| < ϵ holds. So what else do I have to do? Do I have to prove that there exists a δ such that δ=ϵ/3 for all ϵ? Isn't that necessarily true anyway since ϵ > 0 by definition? What do you mean by an epsilon value's being "valid?" What does it mean to prove that a delta is true?

Sorry I'm so dense.

These are one of the hardest parts of calc, so don't feel bad.

The pre-delta/epsilon definition of a limit of the form $$\lim_{x->a}F(x)=L$$ is:

F(x) is arbitrarily close to L for any x sufficiently close to a. The arbitrarily close part is |F(x)-L|<ϵ and it is arbitrary because we define epsilon, or the distance between f(x) and the actual limit. |x-a|<delta is the sufficiently close part, and it is not arbitrary. If epsilon=1, then there exists a delta for which any input "x" that makes |x-a|<delta true will put the function output within one unit (epsilon=1) of the actual limit at a.

In your case, delta=epsilon/3. If I want f(x) to be within one unit of L (epsilon=1), then delta=1/3, which means |x-2|<1/3. In other words, any value of x between 5/3 and 7/3 will put me within one unit of 7.

There are many approaches to proofs, but in this case, the idea was first to find a value of delta by expanding |F(x)-L|<ϵ and performing "cosmetic surgery" to make it look like
|x-a|<δ. From there, you successfully worked backwards and proved that that value of delta made the "epsilon inequality" (i.e. |F(x)-L|<ϵ) true.

Edit: Once you've mastered that, there is one thing that I should add: often times we are faced with either functions that do not have a uniform slope or with arithmetic operations of functions (e.g. find the limit of f(x)+g(x)). This can be an issue as the value of delta that puts f(x) within epsilon of the limit might be different from the value of delta that puts g(x) within epsilon of the limit. In these situations, we often have multiple values for delta, so we take whichever is smallest, which is written as delta=min{value of delta 1, value of delta 2, etc.}. This is important as, in these situations, putting in a value of "x" within delta of "a" will often put you closer to the limit than epsilon required. This isn't relevant to your equation as it was linear and, thus, had a constant slope; however, it is something that you should be aware of.

 

[Nidhogg]

Wow, thanks for all your help, Ciubba! I'll be studying some of the more complex delta-epsilon proofs with a friend this evening and your post will definitely be referenced for that.

 

[ciubba]

If you decide to do general cases, a useful trick is to prove that as epsilon becomes small, so do the "bounds" of the inequalities. Good luck!