Welcome to Calculus. I'm Professor Greist. We're about to begin Lecture 7 on Limits. >> In many respects, calculus can be defined as the mathematics of limits. In this lesson we'll review the concept definition of a limit. Consider a few examples and see that one of the most effective tools for computing a limit involves Taylor series. >> In your previous exposure to calculus, you have certainly seen limits. But what does it mean to say the limit, as x approaches a of f of x equals L. Well, I'm sure you have an image in your head, that as x gets closer and closer to a, f of x gets closer and closer to L. Perhaps you remember that it doesn't matter whether you approach from the left Left or from the right. Perhaps you remember that it doesn't matter what the actual value of the function is at x equals a. What matters is the limit. Well this picture is the intuition behind the limit, but it is not the definition. The definition is another thing all together. What is it? The limit of f of x, as x goes to a, equals L. If, and only if, for every epsilon greater than 0 there exists some delta greater than 0. Such that, whenever x not equal a is within delta of a, then f of x is within epsilon of L. That is a bit of a mouthful, and a lot of students have difficulty with it. Why? Because as a logical statement, it is complex. As a grammatical statement, it is complex. How does one make sense of this? Well, a picture is not a bad way to go. If we think of L as a target that we are trying to hit, then we are allowed some tolerance on the output. This tolerance is in the form of an epsilon. You have to get within epsilon of L. Using your function, you can set the input to be as close to a as you like. But there's going to be some tolerance on the input. Some degree of error that is bounded by delta. In order to have the limit of f of x as x approaches a equals L, then anything within the input tolerance has to hit the target within the output tolerance. 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. The question of evaluating or computing a limit is subtle. And it weaves its way throughout this course. In our next lesson, we're going to consider one of the primary tools for evaluating limits. That of L'Hopital's rule. [BLANK_AUDIO].