Limits

We'll begin by taking an in-depth look at the epsilon delta definition of the limit of a function.

We'll also make several comments in order to clarify the definition. We'll also visit the negation of the definition. What does it mean for a function to not have a particular limit? And we'll finish with two epsilon delta proofs, proving that the limit of a function is a particular value, so you can really start to grapple with the definition.

It'll give you a practice exercise for the road. Before we see the formal definition, let's make sure we're on the same page about what we are trying to capture. The (lim_x→c)(ƒ(x))=L. What should this mean? What should the definition of a functional limit be trying to capture? Well, what it should mean is that as x gets arbitrarily close to c, right, as x approaches c, f of x gets arbitrarily close to L. The limit of f of x as x approaches c is L. This is what the definition should mean.

As x gets close to c, f of x gets close to L. So this is the idea that we are trying to formalize with the definition of the limit of a function. Note that nothing we're saying here involves the actual value of the function at c. We don't care what f of c is. It might not even exist.

We're just trying to define what it means for f of x to approach something as x approaches c. All right, let's see the definition. Let f be a function from a subset of the reals a to the real numbers.

And let c be a limit point of the domain a. Then we say the limit of f of x as x approaches c equals L if for all epsilon greater than zero there exists some delta greater than zero such that for every domain element x that is within delta of the limit point c, we have that f of x is within epsilon of that limit L. And in this case, we say that the limit of f of x as x approaches c converges to L. We think of epsilon as a sort of tolerance, a allowed distance from the limit. And we think of delta as a closeness to the limit point c. When we define the limit of a sequence, we said that for any tolerance, we should be able to go sufficiently far out in the sequence to guarantee that the terms were within the given tolerance of the limit. For a function, we're saying that for any tolerance, we should be able to get sufficiently close to the limit point c so that the function is within the tolerance of the limit.

We could also restate this definition in terms of neighborhoods. This inequality here is describing a delta neighborhood of c with the restriction also that x is not equal to c. The distance has to be positive. And this inequality describes an epsilon neighborhood of the limit L. So we could restate the definition as follows.

The limit of f of x as x approaches c equals L if for all epsilon greater than zero, there exists some delta neighborhood of c such that for all domain elements x in that delta neighborhood, f of x is in the epsilon neighborhood of the limit L. There are several more comments to be made about this definition, but I just want to quickly show you a picture kind of representing the definition. Here we see our function has a limit of L as x approaches c. We could consider some epsilon greater than zero, which gives us this tolerance around the limit. And as long as we are within delta of c, as long as x is within delta of c, we see that our function does stay within epsilon of the limit.

If we reduce the tolerance, a smaller delta will typically be required in order to satisfy the condition, although that's not totally clear in this picture, but that's typically the case. If we want to reduce epsilon and require that the function remain even closer to the limit L, we typically need to require that x stay closer to c and thus reduce delta. But remember, there doesn't have to be a single delta that always works.

There just simply has to be a delta for each epsilon greater than zero. And I think you'll find the importance of this picture more clear once we look at the negation of the definition. But before we do that, let's make a few more comments about this definition.

First, I want to reiterate that c does not need to be in F's domain. Since we need x to approach c, we simply need c to be a limit point of the domain. So that's why we say in the definition that c is a limit point of that domain A. It doesn't actually need to be in the domain.

What f of c is does not matter to us. It doesn't need to be defined. We just need x to be able to approach c. So c needs to be a limit point.

Obviously, the limit as x approaches c doesn't mean anything if x can't approach c. I'm sure you remember from calculus how in a situation like this, the limit of the function as x approaches c is not going to change if we have a hole or if we fill it in with some other value. It doesn't matter at all. We're asking what does the function do as x approaches c. Secondly, in the definition, we mentioned that x is an element of A because, of course, x gets plugged in to F. So we need x to be in the domain of F. x needs to be in the domain of F and x needs to have some distance from c, but a distance that's smaller than delta.

And again, we require that the absolute value of x minus c is greater than zero because we don't care what happens at x equals c. We're not investigating that. And indeed, this is just an economical way of saying that x is not equal to c. Finally, for convenience, we may at times not write these conditions that x is an element of A and that zero is less than the absolute value of x minus c. When we write f of x, it's kind of implied that x is in the domain of F. And when we're talking about the limit of a function as x approaches c, we understand that what happens when x equals c is not relevant. And so we may not always write strictly that x has a non-zero distance from c. But these are important.

Now, let's quickly talk about the negation of the definition, what it means for a function to not have a particular limit. To show that the limit of a function does not equal something, we must prove the negation of the definition. Now, the definition is stated again here, and when we negate the definition, this is what we get.

For all epsilon becomes there exists epsilon. There exists delta becomes for every delta. For every x in A becomes there exists x in A. And f of x minus L is less than epsilon becomes f of x minus L is at least epsilon.

In total, the negation states that there exists some tolerance, epsilon greater than zero, so that no matter how close we force x to be to c, there will be some x which is that close, but where f of x is not within the tolerance epsilon of the limit. The distance between f of x and the limit is greater than or equal to epsilon. That's what it means for this to in fact not be the limit.

Here is a pictorial representation of the negation of the definition. Here, our blue function f of x does not have a limit of L as x approaches c. If we had a bigger epsilon, a wider tolerance, then it would not be evident that the function fails. However, there only needs to exist some epsilon for which the definition fails.

For this particular epsilon, it doesn't matter how close we force x to be to the limit point c. Our function is still going to jump outside the given tolerance. We can make delta as small as we want and get as close to c as we desire. The function's distance from L is still going to exceed epsilon.

So there's an in-depth look at the definition and the negation of the definition. Let's finish off with a few basic exercises proving that the limit of a function is a particular value using the epsilon delta definition. For this first exercise, we're considering the function f of x equals 4x plus 1, simple linear function.

We want to prove that the limit of f of x as x approaches 3 equals 13. We can see that obviously this is true. If you plug 3 into this function, you get 13.

But how do we prove this limit using the formal definition? Well, remember, in the definition of the limit of a function, there needs to exist some delta satisfying the conditions. So we get to pick delta. We can make delta as small as we want.

We just need to make it sufficiently small so that the function stays sufficiently close to the limit. So often for these proofs, you're going to have to go through a phase of scratch work first before writing the proof. Since we can make x as close to c as we want, we can control it by choosing an appropriate value for delta.

So the process we take in the scratch work to figure out how we will write the proof is to assume our desired conclusion that the function is within epsilon of the alleged limit l and to unwind this inequality in order to find a suitable value for delta by solving for the absolute value of x minus c. We can make that as small as we want by choosing the right delta. So this scratch work is the process of figuring out how small we need to make this absolute value in order to get our desired conclusion. And here is that work.

We begin by assuming that our function is sufficiently close to the limit. Our function is 4x plus 1. The limit is 13. And then we can do 1 minus 13 to get that the absolute value of 4x minus 12 is less than epsilon.

Pull out a 4, and we have that 4 times the absolute value of x minus 3 is less than epsilon. And then dividing by 4, we've now solved for the absolute value of x minus 3, which we can control. The absolute value of x minus 3 is less than epsilon over 4. So if we let delta equal epsilon over 4, then x minus c, this absolute value, being less than delta is going to allow us to go backwards through this string of work and reach our desired conclusion.

In our proof, we obviously can't start with a conclusion, but we can take delta to be as small as we want, and through this work, we've figured out a sufficiently small delta. Let's see the proof. Again, this is the formal epsilon delta proof that the limit of our function, the function 4x plus 1, has a limit of 13 as x approaches 3. And here's the proof based on our scratch work.

We begin with an arbitrary epsilon greater than zero. We set delta equal to epsilon over 4. Notice that outside of the context of our scratch work, this value for delta kind of comes out of nowhere. However, we did the work, we know where it comes from, we know that it's all going to work out.

Then, for all x satisfying this inequality, that x is within delta of 3 and their distance is positive, we have that the distance between our function and the limit L is the absolute value of 4x plus 1 minus 13, which goes down to the absolute value of 4x minus 12, pull out a 4, and then this is less than 4 times delta, because the absolute value of x minus 3 is less than delta. We're allowed to make x as close to 3 as we want, because 3 is a limit point of the domain. The domain in this case is just the real numbers.

So since x is within delta of 3, this is less than 4 delta. But delta is equal to epsilon over 4. So 4 delta is equal to 4 times epsilon over 4, which is equal to epsilon. So what we've done is shown that as long as x is within delta of the limit point 3, the distance between our function and the alleged limit 13 is less than epsilon.

Since we've shown that our function is within epsilon of L, for an arbitrary epsilon, the limit of f of x as x approaches 3 is 13. We've proven it. Let's throw down the tombstone and move on to the second practice exercise.

Let f of x equal x squared. We want to prove that the limit of f of x as x approaches 2 equals 4. This problem has an additional wrinkle that's fairly common in these types of proofs. This is all the same stuff we said before.

We can control the absolute value of x minus c. We can make x as close to c as we want, because c is a limit point of the domain, in this case the real numbers. And we can make this absolute value as small as we want by choosing delta. So we need to do some scratch work so that we're choosing a sufficiently small delta.

We assume the absolute value of f of x minus the limit is less than epsilon. In this case, f of x is x squared, and our desired limit is 4. We can factor this.

It's a difference of squares. So the absolute value of x minus 2 times x plus 2 is less than epsilon. The absolute value of a product is equal to the product of the absolute values.

So this means the absolute value of x minus 2, which we can control, times the absolute value of x plus 2 is less than epsilon. This is pretty good. We've almost solved for the absolute value of x minus 2. We can make that as small as we want by choosing an appropriate delta.

But what do we do about the absolute value of x plus 2? Can we say anything about that? Well, yes we can, because we can make x as close to 2 as we want. For example, we could force x to be within 1 of 2, in which case x could be no more than 3. But that puts an upper bound on the absolute value of x plus 2. In that case, the absolute value of x plus 2 could be no more than 5. And here that is written out. Since x is close to 2, x plus 2 can only be so big.

Let's say delta is less than or equal to 1. Any value would work. We just got to pick one. Let's say we pick 1. Then x being within delta of 2 implies that x is within 1 of 2, which means that x is between 0 and 3. But that means the absolute value of x plus 2 must be less than 5. Alright, we've got an upper bound, so we can proceed with the inequality.

We had that the absolute value of x minus 2 times the absolute value of x plus 2 was less than epsilon. But now we know, given what we've chosen for delta, that it has to be less than or equal to 1. We know that x plus 2 is less than 5. So the absolute value of x minus 2 times the absolute value of x plus 2 must be less than the absolute value of x minus 2 times 5. Because again, the absolute value of x plus 2 is less than 5. And then this is less than epsilon. And we can divide both sides by 5 in order to get this additional restriction.

The absolute value of x minus 2 must be less than epsilon over 5. So this is another condition we need to satisfy. We need delta not only to be less than or equal to 1, so that we have this bound on x plus 2, but we also need delta to be less than or equal to epsilon over 5. We can accomplish this, satisfying both conditions, by setting delta equal to the minimum of those two requirements. The minimum of 1 and epsilon over 5. That way it's guaranteed that delta is less than or equal to 1 and less than or equal to epsilon over 5. Because whichever one is smallest, that's what delta is going to equal.

This is pretty common, that we'll need to put multiple restrictions on delta, and we can do so with a minimum function. And now we can do the proof. Let epsilon be greater than 0, and set delta equal to the minimum of 1 and epsilon over 5. Then for all x that are within delta of the limit point 2, we have, as previously discussed, that x is less than 3, because it has to be within 1 of 2. So x is less than 3. And thus we have that the absolute value of f of x minus L is equal to the absolute value of x squared minus 4. Factoring the difference of squares, we have the absolute value of x minus 2 times x plus 2. We can split up the absolute value of a product, so we have the absolute value of x minus 2 times the absolute value of x plus 2. But we said that x is less than 3. We could also throw in that it's greater than 0. And so the absolute value of x plus 2 is less than 5. We already went over that.

That means that this, since the absolute value of x plus 2 is less than 5, this is less than this. But the absolute value of x minus 2 is less than epsilon over 5, because it's less than delta. And delta is the minimum of 1 and epsilon over 5. So for sure, the absolute value of x minus 2 is less than epsilon over 5. So this is less than this.

And this equals epsilon. Thus, for an arbitrary epsilon, we've shown that if you make x sufficiently close to c, then the absolute value of f of x minus L is less than epsilon. Thus, by definition, the limit of f of x as x approaches 2 is equal to 4. I'll just scroll through to give you one more look at the scratch work and the proof in its entirety, and scroll up to the first problem and do the same thing.

So those are two basic examples of using the epsilon delta definition to prove that the limit of a function at a particular point is a particular value. Again, the definition simply says that the limit of f of x as x approaches a limit point c equals L if for every epsilon greater than 0 there is some delta neighborhood of the limit point c, so that for all domain elements x that are within delta of c, we have that f of x is in that epsilon neighborhood of the limit L. That's what it means for the limit to converge to L. Now that we understand the definition of a functional limit, we'll be able to prove some of its basic properties and finally go on to define continuity once and for all. Here's one more practice exercise for you to try yourself.