Strategy in finding limits | Limits and continuity | AP Calculus AB | Khan Academy

Multiple videos and exercises we cover the various techniques for finding limits, but sometimes it's helpful to think about strategies for determining which technique to use, and that's what we're going to cover in this video. What you see here is a flowchart developed by the team at Khan Academy, and I'm essentially going to work through that flowchart. It looks a little bit complicated at first, but hopefully, it'll make sense as we talk it through.

So, the goal is, hey, we want to find the limit of f of x as x approaches a. So, what this is telling us to do is, well, the first thing, just try to substitute. What happens when x equals a? Let's evaluate f of a. This flowchart says, well, if f of a is equal to a real number, it's saying we're done. But then there's this little caveat here, probably. And the reason why is that the limit is a different thing than the value of the function. Sometimes they happen to be the same; in fact, that's the definition of a continuous function, which we talk about in previous videos.

But sometimes they aren't the same. This will not necessarily be true if you're dealing with some function that has a point discontinuity, like that, or a jump discontinuity, or a function that looks like this. This would not necessarily be the case. But if, at that point, you're trying to find the limit towards the if, as you approach this point right over here, the function is continuous, it's behaving somewhat normally, then this is a good thing to keep in mind. You could just see, hey, can I just evaluate the function at that, at that a over there?

So, in general, if you're dealing with pretty plain vanilla functions like an x^s, or if you're dealing with rational expressions like this, or trigonometric expressions, and if you're able to just evaluate the function, and it gives you a real number, you are probably done. If you're dealing with some type of a function that has all sorts of special cases and it's piecewise defined, as we've seen in previous other videos, I would be a little bit more skeptical. Or if you know visually around that point there's some type of jump or some type of discontinuity, you've got to be a little bit more careful. But in general, this is a pretty good rule of thumb. If you're dealing with plain vanilla functions that are continuous, if you evaluate at x equals a and you get a real number, that's probably going to be the limit.

But now let's think about the other scenarios. What happens if you evaluate it and you get some number divided by zero? Well, that case, you are probably dealing with a vertical asymptote. And what do we mean by vertical asymptote? Well, look at this example right over here, where you're saying the limit that in a darker color. So if we're talking about the limit as x approaches 1 of 1 over x - 1, if you just try to evaluate this expression at x = 1, you would get 1/1 - 1, which is equal to 1 over zero, which says, okay, I'm throwing it; I'm falling into this vertical asymptote case.

And at that point, if you wanted to just understand what was going on there or even verify that it's a vertical asymptote, well then you could try out some numbers. You could try to plot it. You could say, all right, I probably have a vertical asymptote here at x = 1, so that's my vertical asymptote. You could try out some values. Well, let's see, if x is greater than 1, the denominator is going to be positive, and so my graph, and you would get this from trying out a bunch of values, might look something like this. And then for values less than 1, or less than one I should say, you're going to get negative values, and so your graph might look like something like that. And so you have this vertical asymptote; that's probably what you have.

Now, there are cases, very special cases, where you won't necessarily have the vertical asymptote. One example of that would be something like 1/x - x. This one here is actually undefined for any x you give it, so it would be very, you will not have a vertical asymptote, but this is a very special case. Most times, you do have a vertical asymptote there. But let's say we don't fall into either of those situations. What if, when we evaluate the function, we get 0 over 0?

And here is an example of that limit as x approaches -1 of this rational expression. And let's try to evaluate it. You get -1^2 which is 1 plus -1 which is +1, minus 2, so you get zero in the numerator, and then in the denominator, you have 1^2 which is 1, minus 2 * 1, so +2 minus 3, which is equal to 0. Now this is known as indeterminate form, and so on our flowchart we then continue to the right side of it.

And so here's a bunch of techniques for trying to tackle something in indeterminate form. And likely, in a few weeks, you will learn another technique that involves a little bit more calculus called L'Hôpital's Rule that we don't tackle here because that involves calculus, while all of these techniques can be done with things before calculus, some algebraic techniques and some trigonometric techniques.

So the first thing that you might want to try to do, especially if you're dealing with a rational expression like this and you're getting indeterminate form, is try to factor it. Try to see if you can simplify this expression. And this expression here, you can factor it. This is the same thing as x - 2 * x + 1 over x - 3 * x + 1. If what I just did seems completely foreign to you, I encourage you to watch the videos on factoring polynomials or factoring quadratics.

And so you can see here, all right, look, if I make the C, I can simplify this because as long as x does not equal -1, these two things are going to cancel out. So, I can say that this is going to be equal to x - 2 over x - 3 for x does not equal 1. I sometimes people forget to do this part; this is if you're really being mathematically precise. This entire expression is the same as this one because this entire expression is still not defined at x = 1. Although you can substitute x = 1 here and now get a value, so if you substitute x = 1 here, even if it's formally, if we're formally taking it away to be mathematically equivalent, this would be -1 minus 2 which would be -3 over -1 - 3 which would be -4, which is equal to 3/4s.

So if this condition wasn't here, you could just evaluate it straight up, and this is a pretty plain vanilla function. Wouldn't expect to see anything crazy happening here, and if I can just evaluate it at x = 1, I feel pretty good. I feel pretty good.

So once again, we're now going into factoring. We were able to factor, we evaluate, we simplify it, we evaluate the expression, the simplified expression now, and now we were able to get a value. We were able to get 3/4s, and so we can feel pretty good that the limit here, in this situation, is 3/4s. Now let's categorize what we've seen so far as the bulk of the limit exercises that you will likely encounter.

Now, the next two I would call slightly fancier techniques. So if you get indeterminate form, especially, you'll sometimes see it with radical expressions like this rational radical expressions. You might want to multiply by a conjugate. So, for example, in this situation right here, if you just tried to evaluate it x = 4, you get the square root of 4 - 2 over 4 - 4, which is 0 over 0. So it's that indeterminate form, and the technique here, because we're seeing this radical and a rational expression, is hey, let's maybe, maybe we can somehow get rid of that radical or simplify it somehow.

So let me rewrite it x - 2 over x - 4. We say a conjugate, let's multiply it by the square root of x + 2 over the square root of x + 2. Once again, I'm, it's the same expression over the same expression, so I'm not fundamentally changing its value. And so this is going to be equal to, well, if I have a + b * a minus b, I'm going to get a difference of squares. So it's going to be square root of x^2 - 4 over, well, square root of x is just going to be x - 4.

So let me rewrite it that way. So it's x - 4 over x - 4 * square root of x + 2. Well, this was useful because now I can cancel out x = 4 or x ≠ 4 right over here. And once again, if I wanted it mathematically to be the exact same expression, I would say, well, now this is going to be equal to 1 over the square root of x + 2 for x does not equal 4. But we can definitely see what this function is approaching if we just now substitute x = 4 into this simplified expression, and so that's just going to be one over, so if we just substitute x = 4 here, you'd get 1/square root of 4 + 2, which is equal to 1/4.

And once again, you can feel pretty good that this is going to be your limit. We've gone back into the green zone. If you were actually to plot this original function, you would have a point discontinuity; you would have a gap at x equals four, but then when you do that simplification and factoring out that x or canceling out that x = 4, that gap would disappear. And so that's essentially what you're doing; you're trying to find the limit as we approach that gap, which we got already.

Now this final one, this is dealing with trig identities, and in order to do these, you have to be pretty adept at your trig identities. So if we're saying the limit as, let me do that in a darker color. So if we're saying the limit as x approaches zero of sin of x over sin of 2x, well, since 0/0, you're going to get 0 over 0 once again, indeterminate form. We fall into this category, and now you might recognize this is going to be equal to the limit as x approaches zero of sin of x. We can rewrite sin of 2x as 2 sin x cosine x, and then those two can cancel out for all x's not equaling, uh, for all x's not equaling zero, if you want to be really mathematically precise.

And so there would have been a gap there for sure on the original graph if you were to graph y equals this. But now, for the limit purposes, you could say this limit is going to be the limit as x approaches zero of 1/2 cosine of x. And now we can go back to this green condition right over here because we can evaluate this at x equal 0; it's going to be 1/2 * cosine of 0. Cosine of 0 is 1, so this is going to be equal to 1/2.

Now, in general, if none of these techniques work and you'll encounter a few other techniques further on, once you learn more calculus, then you fall on the baseline approximation, and approximation you can do it numerically. Try values really, really, really, really close to the number you're trying to find the limit on. You know, if you're trying to find the limit as x approaches zero, try 0.00001, try negative 0.001. If you're trying to find the limit as x approaches four, try 4.1, try 3.99999999, and see what happens. But that's kind of the last ditch, the last ditch effort.