Hello!
Apologies for the long delays in updating this journal. I've had a lot of things happening, not all of which have been entirely productive. Also this is the first conversational calculus entry typed from an offline livejournal client running under my new Ubuntu system. Well, the system itself is about a year old, but I am finally running Ubuntu on it properly (which is what I bought it to do).

Now in previous posts I was talking about limits. Well, everything that I was talking about had to do with having a general idea of how numbers and equations work to work out what the limit is. Today I'm going to talk about proving limits.

If you are in a full-on general calculus course then there is probably at least half a chance that you will get into this topic. If you are in a business or applied calculus course then it probably won't come up. Truthfully, I'd like to see it left until an advanced level course in calculus because I think that it's unclear what is wanted and can confuse students to no end.

If you have been in a high school geometry course, you probably have has some exposure to proofs. In such a course, however, you are usually not using formal rules, but instead sort of showing why something works in clear language. Well, delta-epsilon is a kind of proof. The good news, though, is unlike geometry where you had to really work out how to prove something, all of the delta-epsilon proofs have a simple structure that you have to follow.

To begin with, this is a lowercase delta - δ
and this is a lowercase epsilon - ε

I think that I have seen a textbook somewhere that used non-Greek letters for δ and ε, but I'll stick to the Greek for now.

Now let's look at a relatively simple statement of limits:


Nothing tricky here, right? Just plug the 3 in for x and you get 7. Nothing funny. Here is the statement that I'm going to use to prove that:


WOWZERS! What the funk does that mean?

Well there is a lot of symbolic mathematical notation used above, so I'll do over it. Your teacher might accept your solution written without all of the symbolic stuff, but it would probably impress him or her if you casually knew how to use it.

means 'for all.' It's just an A flipped over.
'numbers epsilon greater then zero'. You should know that one.
means 'there exists.' This flipped over E was probably crafted by the same people who made the flipped over A.
'some number delta greater then zero.' I changed the wording slightly from the inequality above in order to have the sentence make sense.

So far we have 'For all numbers ε greater then 0, there exists some number δ greater then zero.'

Well, that's true, right? If you say 'Epsilon is 4, and epsilon is greater then 0,' then I can say 'Delta is 5 (or 6, or 3, et cetera) and delta is greater then zero.' We are not really saying anything so far. All that we are saying is that if one number greater then zero exists, then so does another.

But it's not so simple. Let's finish reading the above sentence.

'such that' (easy to remember, no?)
'0 is less then the absolute value of x-3, which is less then δ' which essentially is saying that the distance between our number x, and the limit that x is approaching (which is in this case 3) is less then our δ from the beginning (delta is usually presumed to be quite small)
'means that' or 'tells us that'
'the absolute value of the difference between our function (2x+1) and the number that we claim as our limit (7) is less then &epsilon'.

Let's take this from the top.

'For all numbers ε greater then 0, there exists some number δ greater then zero such that if 0 is less then the absolute value of x-3, which is less then δ then the absolute value of the difference between our function (2x+1) and the number that we claim as our limit (7) is less then &epsilon.'

or as you would actually read it out loud:

'For all numbers ε greater then 0, there exists some number δ greater then zero such that if 0 is less then the absolute value of x-3, is less then δ then the absolute value of (2x+1)-7 is less then &epsilon.'

So what's the big idea? Well, let's pretend that I thought that I thought that was not 7. So I said, 'Hey, smarty-pants! This function might come close to 7 as x comes close to 3, but it doesn't come that close. You'll never get it closer then .1 away from 7 (in the range of 6.9 to 7.1).' They you might say, 'Look, booger-head, as long as δ (the distance between x and 3) is less then .05, then the function comes to within .1 of 7.'

So I whip out my calculator to check your claim. Well, if I plug in 3+.05=3.05 into the function then I get , which is .1 away from 7. If I plug in 3-.05=2.95, then I get , which is also .1 away from 7. If I choose some δ less then .05, say .04, then I might get , which is less then .1 from 7.

So I say, 'Curses, vile fiend! OK, you bested me once, but can you do it again? There is no way that you can get within .03 of 7.' You turn around and say, 'Just make sure that your x is within .015 of 3.' If you try this in your calculator, then you will notice that you have won again.

Well, I can keep throwing challenges at you, but the only way to prove that is to tell me your secret for picking δ to beat my ε. And that is what the actual proof consists of.

Annnd I'm going to call it quits for tonight and finish this hopefully tomorrow. It will be easier to pick it up if I know that I am half-finished.

OK, this is the third post on limits and it is my sincere hope to finish off that topic, and also the topic of continuity, so that we can get into the good stuff. Different classes will spend different amounts of time on limits and continuity but I want to try to hit as many points as possible so that this resource can help as many people as possible.

To begin with, do you remember the diagram from the last post? It occurs to me that in some textbooks, the diagram might look more like how you see it below:



See those little empty or filled in holes? They tell you whether the function is or is not defined at those points. Think of them as the graphical equivalent of using either , signs, or , signs. If you see an open loop, such as at (5,5), then it means that that point (5,5) does not exist on the graph. In other words, . In this case, f(5) takes on no value. However, if you see a filled loop, such as at (-2,-2), then it means that the function is defined at that point.

So, see if you can tell me the value of f(8).

( Solution here. )

I’m going to move on to discuss the idea of continuity before looking at how to find limits algebraically.

Think back to the first lesson when our friend Zack was strolling along the function curve, and we talked about what would happen if unbeknownst to him there was a hole in the function for him to trip over. Mathematically, we would say that the function has a discontinuity at that point. As far as we can see, the function was continuous at all other points. Let’s look at an example from the first post:



This function is discontinuous at point a, but continuous at all over visible points.


Here is the formal definition of continuity at a point a, should you ever have to invoke it:

1. exists.
2. exists.
3.

In so many words, it says that the limit of a function at point a is the same as the value that the function takes on. If you were riding your bike you wouldn't have to worry about potholes or jumping a kerb or anything.

Do take note, however, that in the first example in this post, that were the hole filled in at (5,5) then the function would be continuous at that point. Even though there would be a sharp angle in the function curve, it would still be true that (1) f(5) exists, (2) exists, and (3) .

Now I'm going to look at some methods for analysing limits algebraically.

What happens if we look at a plot of the function (Technically since the coordinates are defined as x and y, not x and f(x), we need to set y equal to the value of the function to plot it. For purposes of your class you could also think of this as being simply either or .)

Well, my handy-dandy TI-84 gives me this picture:


OK, uhmmmm, it looks like a normal line but . . . there's a hole in it? How did we get something like that out of a rather difficult to analyse looking function such as ?

Well, here is how we deal with it. Notice that the denominator (bottom of the fraction) is . This means that when we get to the point , we will be dividing my zero. As you should know by now, if you try to divide by zero then your homework bursts into flames, your calculator explodes into billions of tiny pieces, and so on. So, to algebraically find the limit at x=5, we have to somehow get that nasty out of the denominator, in order that the universe may again be at peace.

Well, let's see if we can factor an out of the numerator (the top of the fraction). It so happens that we can factor the numerator into . Thus, the function becomes . Now we can pull a out of the numerator and denominator and simply get . Since we can safely plug 5 into this function and get , we can say also of the function in it's original form that .

So, If the denominator goes to zero at a point, see if the denominator is a factor of the numerator, and if so then factor it out and get rid of it.

There are two more forms of limits that can be found algebraically that I can't find examples of in either of the textbooks that I am working with. I will make a note to look for examples elsewhere and will probably append this entry later.

For now, I am going to talk about the exciting world of the δ-ε (delta-epsilon) proof of a limit.

I wish that I could avoid this topic. If you are taking a business or applied calc course then you can probably skip this section. In normal calc classes there is some chance that it may come up, and if it does then it's confusing enough to deserve some mention here.

Annnnd, I'm going to save it for tomorrow (or possibly Thursday, depending how my week goes) because I've been typing this for quite some time and I have other things to do today.