This must be true no matter how small the output tolerance epsilon is.

You can find some sufficiently small input tolerance to guarantee always

striking within range of the limit. In the context of an actual function f,

one can visualize this delta epsilon definition as follows.

You choose an output tolerance, epsilon. Then, there must be some input tolerance,

delta, so that any input within delta of a has an output within epsilon of L.

Now, many students get confounded here, trying to find the optimal delta.

It does not need to be optimal. You can choose something smaller, that is

not a problem. The critical part of the definition is,

that as you change epsilon, you need to be able to update delta.

If you make epsilon smaller still and decrease your level of acceptable error

on the output, you need to find some amount of acceptable error on the input.

And this has to continue for every possible non-zero value of epsilon.

That is what captures what the limit is. This view of the definition is extendable

to other context. Consider the limit as x goes to infinity

of f of x. What does it mean for that to be equal to

L? Well, we're going to think about infinity

as something like an end point to the real line, modifying its topology so that

it looks like a closed interval. Now, this is a dangerous thing to do if

you don't know what you're doing. But let's think about it from the

perspective of our interpretation of a limit.

Given any output tolerance, epsilon, there must be some tolerance on the input

that guarantees striking within epsilon of L.

Now, how do we take a neighborhood of infinity?

How do we talk about a tolerance on that input.

Well, what it becomes in this context is sum lower bound M, so that, whenever your

input is greater than M, then your output is within epsilon of L.

As before, this must be true no matter what epsilon you choose.

If you make your tolerances on the output, tighter and tighter, then, we can

make the tolerances on the input tighter and tighter.

In this case, instead of talking about being within delta of infinity, since we

are only looking at a one sided limit. We can speak in terms of an explicit

lower bound on x, the same intuition and picture holds.

To be sure, not all limits exist. Not all functions, well behaved.

There are several ways in which things can go wrong.

You could have a discontinuity at a function.

The limit would not exist at that point. You could have what is called a blow up,

that is the function goes to infinity as x gets closer and closer to a.

Or, worse still, the limit can fail to exist because of an oscillation, where

the function oscillates so badly that the limit at a does not exist.

On the other hand, most of the time you're not going to have to worry about

this because most functions are continuous.

And we say that f is continuous at an input a, if the limit, as x goes to a, of

f of x exists and equals f at a. We say that f is continuous everywhere.

If this statement is true for all inputs a in the domain.

Now most of the functions that we're used to seeing are continuous functions and

one doesn't have to worry so much about limits in this case.

There's a bit of a technicality. One has to be very explicit about which

points are in the domain of the function. Some functions which look discontinuous

may actually be continuous. If the discontinuous looking point is not

actually in the domain, if however, the function is defined there, then, the

discontinuity presents itself. There are certain rules associated with

limits that you may know by hand, even if you don't remember them.

If the limit of f of x, as x approaches a, and the limit of g of x, as x

approaches a, both exist, then the following rules are in effect.

There's a summation rule that the limit of the sum of f plus g is, in fact, the

sum of the limits. There is, likewise, a product rule, that

the limit of the product of f and g is, in fact, the product of the limits.

There is, likewise, a quotient rule, that the limit of f divided by g is the limit

of f divided by the limit of g. Now, at this point, you've gotta be a bit

careful if that denominator is zero. Well, then this limit may not exist.

There's, likewise, a chain rule or a composition rule that says the limit of f

composed with g, as x go to a, can be realized as f of the limit of g as x

approaches a. Now once again, this too has some

conditions. f, in this case, needs to be continuous

at the appropriate point in order for this to hold.

Now, at this point, I think we're going to have a little quiz to test your

knowledge of limits. What is the limit, as x approaches zero,

of sine of x over x? And this is a quotient.

Can we apply our quotient rule for limits?

No, I'm afraid we cannot because the denominator is going to 0 and so is the

numerator and 0 over 0 presents some difficulties.

Now, I bet that most of you know the answer is 1 but why do you know this?

Well, you may say, I remember this. This is something that I had to memorize

when I took high school calculus very useful on exams, I just know it.

Well, that's not a very satisfying answer is it?

Some of you may say, I wield the mighty sword of L'Hopital's Rule and I know that

if I differentiate the top and the bottom, then I get one.

That's great, and I'm glad you remember L'hopital's rule.

But do you know why it works? Do you have a good reason for your belief

in this rule? Well, if not, then let's take a method

that we do trust. Namely, Taylor series.

If we consider the limit, as x goes to 0 of sine of x over x.

We know what sin of x is. That's x minus x cubed over 3 factorial

plus higher ordered terms. Now, thinking of this, as we do, as a

long polynomial. What are we tempted to do?

Well, I look at that and say hey, we could factor out an x, from the numerator

and cancel that with the x in the denominator.

Yielding the limit as x goes to 0, of 1 minus x squared over 3 factorial plus

higher order terms in x. Sending x to 0 gives us an answer of 1,

and the limit makes perfect sense. Likewise, you might recall, that the

limit as x goes to zero of 1 minus cosine of x over x, is, now what was what?

Oh well, I don't remember, but I do remember what cosine of x is.

And I note that here, the ones cancel. And I'm left with the limit of x squared

over two factorial minus x to the fourth over 4 factorial, plus higher order

terms. When I divide that by x, I get the limit

of x over 2 factorial, minus x cubed over 4 factorial, plus higher ordered terms.

There's no 0 over 0 ambiguity any more. This limit is precisely 0.

Now, one of the wonderful things about this Taylor series approach to limits is

that it works even in cases where you might not have memorized the limit and

where the limit is, indeed, not so obvious.

Well, let's look at the cube root of 1 plus 4x minus 1 over the fifth root of 1

plus 3x minus 1. It is clear that evaluating at zero is

not going to work, that yields 0 over 0. So, what do we do?

Well, rewriting this a little bit allows us to use the binomial series.

With alpha equal one third in the numerator, and one fifth in the

denominator. Applying that gives us 1 plus one third

times 4x plus higher order terms. Subtract 1.

In the denominator 1 plus one fifth times three x plus higher of order terms,

subtract 1. Those subtractions, get rid of the

constant terms, were left with terms that all have an x in them.

We factor that out, and then the leading order terms are four thirds in the

numerator and 3 5ths in the denominator. Yielding an answer of 20 9ths.

That is beautiful. >> There's a vast gulf between knowing

how to compute something and knowing what that thing is.