Epsilon-delta definition of Limit #1

Brief explanation: Today we start with a topic I’ve struggled a lot over the past year and that I still don’t fully understand. This topic is the epsilon-delta definition of a limit. This is a rigorous definition that mathematicians use to prove what the limit of a function is when we approach a certain value. Now, I’m not going to explain the theory behind these proofs, I’m actually planning on solving some exercises (with different difficulties) and posting them here. If you want to understand better the theory behind these proofs I’d advise you to look in these resources:

Today we start with a very simple example that will help you to start to get the grasp of this.

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Question: Prove the following limit:

Enunciado.png

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Comprehending: By now I hope that you know the very basics of epsilon-delta proofs, so I’ll start by stating the sentence that you must know by heart (and obviously, what it means):

For every ε > 0 there is some δ > 0 such that for all x,

if 0 < |x – a|< δ then |f(x) – l|< ε.

(where a is the value the function tends to and l the value of the limit)

This statement tells us everything: there is a relation between our δ and our ε. We have to choose a δ such that the condition |f(x) – l|< ε is implied. In mathematical terms:

epsilon delta.png

Now let’s apply this to our case.

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Solving: First of all we apply the above conditions to our case:

1.png

All we need is to find a δ as a function of ε that implies the conditions given. With a bit of intuition we can write:

2.png

The first step was just intuition that you need to develop, while the second step is a property of absolute values that you need to know: if you have terms multiplying inside the absolute value bars, you can separate them while maintaining the bars on each term.

Now let’s isolate the term we have that is equal to the term on the delta inequality.

3.png

(Notice how both left sides of this and the delta inequalities are the same. We are on the right track!)

The problem here is that we can’t have any x‘s when defining our delta! The trick when dealing with this is: limiting our term related to the δ and predict something about the term under the epsilon. Let’s apply this trick. A common value to apply here is 1, but any value will do! Normally people don’t use big values because we are defining an interval as close as we want, so a big value would not make sense here (but, as far as I know, would prove this as well). Applying this:

4.png

Remember the basic inequality of absolute value |x|-|a|<|x – a|:

5.png

Taking the left and right terms of the inequality:

6.png

Now let’s see what this will tell us about |x+3|. Remember another basic inequality of absolute value |x+a|<|x|+|a|:

7.png

(In the middle we used the value we deduced to be bigger than |x|)

Now, we have the bound we wanted! Returning to our epsilon equation: if |x + 3| < 7, then,

8.png

(I’m actually unsure of this step: I’ll most likely review it)

But remember that to get this value we had to bound |x-3|, so the answer will have to be:

9.png

So if ε/7 is bigger than 1, our delta will be 1! This completes the proof. Now all we need to do is to rewrite the proof assuming the delta values and show how that means our ε will be bigger than |f(x) – l| (I won’t do it here because it would make this post much longer).

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Conclusion: We were able to prove our limit rigorously, thanks to a trick commonly done in these situations. Please, any doubts you have leave in the comments and I’ll try to answer!

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3 thoughts on “Epsilon-delta definition of Limit #1

  1. Pingback: Epsilon-Delta definition of Limit #2 – Matiphy

  2. Pingback: Epsilon-Delta definition of Limit #3 – Matiphy

  3. Pingback: Epsilon-Delta definition of Limit #4 – Matiphy

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