John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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John Postovit
John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.


Limits are the basis of everything we do in calculus and the AP test, it always asks you a few. There's always going to be one or two of them in the multiple choice section, usually in the part where you don't have to use your calculator. Well, we are going to start off by reviewing some of the basic limits. And then we are going to move on to do some of the more complex types of problems that you might see on the AP test. Let's get started.

Basic limits. Let's have a look at few. Basic limits come in 4 fruit flavours. You've got finite numbers divided by infinity, you've got finite numbers divided by 0, you've got infinity divided by infinities, and 0 divided by 0. These are all impossible situations.

How do you divide by infinity? Well, you can't. That's where limits come in. How do you divide by 0? You really can't. So that's why you need limits. Limits are ways of finding numbers that are impossible numbers. You can almost quite get to but really never get there until you're infinitely close. Here is the first kind, finite divided by infinity. This will be kind of tough. 100,000 divided into an infinite number of parts. If you divide 100,000 into one part, you get 100,000. You divide it into 10 parts you get 10,000. If you divide it into 100,000 parts you get 1. So the bigger the number on the bottom gets, the smaller the result gets. But if you divide 100,000 into an infinite number of parts, each of them has to be infinitely small, in other words, 0. So anything that's finite divided by infinity, always going to be a 0. No matter how big it is. This could be 62 billion up here and if you divide it by infinity, it'll still be a zero.

Then you've the opposite, taking something finite and dividing it by 0. So if you started off with saying, we've got x is 1, you divided this by 1, you get -1000. If you make x smaller like 0.5, -1000 divided by 0.5 is -2000. It's going to be bigger. I'm going to do a quick graph of this one. So, you might recognize this as being a basic hyperbola function. Looks something like that and it's asymptotic to the y axis.

Here, we've gotten into a little bit of a problem. What happens is you are coming from the right side, dividing by smaller and smaller positive numbers, the result gets more and more negative. If you come around this way though, dividing by numbers that are negative but they became closer and closer to 0, this one grows more and more positive. That's what the little plus is about. It's about handling that little detail. This says we are going to approach 0, the numbers here are going to get closer and closer to 0 but they are going to come in from the larger side. So you have -1000 divided by smaller and smaller positive numbers. Look where it's going, negative infinity.

We'll look at this one again. Now, we are going to have the thing just approach 0, not going to worry about that positive or negative part. This time we've got a problem if we come in from the right side with a plus there, the result goes towards negative infinity. If we come in from the left side, if there was a negative there, we'd come in and approach positive infinity.

That last example is called a one sided limit. This is a two sided limit which is the normal kind of limits. Problem is, how do you decide you want to go towards negative infinity or positive infinity? You don't get to decide. If the limit isn't the same coming in from both directions, you just can't say that there is a limit at all. You have to say that it does not exist.

Next flavour. Infinity divided by infinity. That's weird. Something divided by itself. That's usually 1. But weird things can happen. Watch this. There is a neat little trick for doing this limit. You probably saw this one first in pre-calc and you probably reviewed it again at the beginning of this year. If you have something that's going towards infinity, and you have powers of x on the top and bottom, look for the largest power of x in the denominator. That's x³. I'm going to divide the top by x³, which I can do legally as long as I divided the bottom by x³. I'm going to do that. Divide the numerator by x³. Divide the denominator by x³. And now, I'll do a little bit of simplifying. See that x³ can divide with the x³, that x can divide with the x³.

Let me write out that simplification. So we've got our limit as x approaches infinity in the numerator. x³ divided by x³ is 1. X divided by x³ is 1/x². Down here, 2x³ divided by x³ is 2 and x² divided by x³ is 1/x. We're in business now. I can actually put the infinity in now because recognize that?

1/x³? Finite divided by infinite. Finite divided by infinite is 0. Doesn't matter any more. And down here, a finite number divided by an infinite number, that's 0. So that pesky dividing by infinity it's gone. What's left? 1/2. Crazy, huh? This whole thing just simplifies to just 1/2. If you were to put this out on graph, that formula, what you'd find is that it levels off at a height of 1/2. You've probably done this before. Horizontal asymptotes. We'll handle those in another episode.

Last type; 0 divided by 0. You can't divide anything by 0 including itself. So we are going to have to do a limit again because, limits let you divide things that can't normally be divided. In this one, the way to get this ready is to factor. Factoring is going to make this come out nice and neat. On the top, factoring out, x²+x-6. Just a plain old binomial factor like you've been doing since algebra 1. Nice and simple. If you were to graph this, one of the things you'd be expected to spot is that you have the x - 2 factor on the top and the bottom. That means it can divide out. But on the graph, when you divided it out, you wind up having a little gap in the graph. No need to worry about that right now though.

All we have to worry about is this. At the beginning we put the 2 in there because that would force us to do 2 - 2 which is 0. You can't divide by 0. But now all of a sudden, you're not dividing any more. All you have to do is put the 2 in for the x and your result is 2 + 3 or 5.

If you graph this formula, as you got closer and closer to 2, you'd have a spot right on the graph at 5, an open circle.

Let's tackle some sample problems. Now, the problems I'm going to show you in this section are ones along the lines of what you are going to see on the AP test. They'll rarely ask any of the basics. The ones I'm going to show you, now are ones that come out very commonly. First off, we've got two really important limits that you need to memorize. One of the limits is sine x/x as x approaches 0, only as x approaches 0 but the result is 1. Kind of surprising because the sine of 0 is 0 and dividing 0 by 0, how does that give you a 1? Well, it does in this case.

There is a proof for why this works. It can't be done with any of the basic techniques I just showed you. The other one that you might see is 1-cosine x/x. The limit of this one as you approach 0 is 0. Well, sorry to tell you this but memorize them, get over it. They are on the AP test. You don't have time to prove them, you don't have time to come up with them from basics. Just memorize them. Especially this one. If you are going to memorize any one limit, this is it. Let's use it in a problem. This is a really common variation that you can see on this. You might think, sine x/x, the limit is going to be 1. There is a little problem, this is 5x and that's 2x. It's not really the same variable.

The key to this one is realizing you have to make it be the same variable. Over here I started doing that. the first thing I did was to factor this 2 out of the denominator. That's to clear the way forward I need. The variable of that the sin is acting on is 5x.

And I'll be okay if the bottom's got the exact same variable on it, 5x. This can be really confusing for students because you know well, that's not one variable, it's 5x. X is a variable, 5x isn't. Well, variables can have multiple components. You can treat 5x as if it's a single variable and that's going to let this part of the limit become 1. Of course, I can't just multiply the bottom by 5 and say it's the same thing, no way. But I can balance it out. I divided the bottom by 5, so if I multiplied the top by 5, then business, I've rebalanced it. We're almost done. I'm going to do this limit.

The 1/2 stays there, the limit of the sine of some variable divided by the same variable as you're approaching 0 is 1. Times the 5 over 1 factor and, we're going to town. Answer is 5 1/2. Let's try another one. This one is tempting. You might say, Oh my gosh, look at this. The square there, the square there, they balance each other out. We can go to town that limit is just 0. No, sorry. The limit for cosines is 1 minus just a plain cosine over just a plain x. But I can do some things with this. The reason I wrote this down here, the Pythagorean identity, is because it's the key to doing this one. I'm going to rearrange this. I need 1 - cosine squared so I could take this and subtract cosine squared from both sides. So I get sin squared equals 1 - cosine squared. I hope you'll forgive me for being lazy, I really shouldn't leave the variable off but I want to save time. 1 - cosine squared can be replaced with sine squared and I'm going to do that now.

1 - cosine squared is sine squared. I'm putting the variable back in because it really does matter now. The variable have to match up. Wait a minute, it's not that cosine one after all. It's really the sine one. It's still not quite like what we need it to be though. Because it's squared on top and squared on bottom, but that's easy enough to handle. Sine squared x is the same thing as sine x times sine x. So I can split this up into two different limits. The limit as x approaches 0 of sine x/x times another sine x/x. And now I've got two of those limits that we just did.

This limit is one, that one's 1. 1 times 1 is 1. Here is another one of a type that you maybe are a little less likely to see but it's been known to show up on the test and it's a really surprising thing. Take a look at this right here. We've got a square root on the top and a number on the top, no square root on the bottom. Remember back in Algebra 1 and especially Algebra 2 and a lot in Pre-calculus, when you were told over and over again, never leave square roots on the bottom. Guess what? We are going to make this square root be on the bottom. It's a neat little trick that's going to make this one work out. Now, what I'm going to do is multiply this by the conjugate of this, top and bottom. So that is square root of 9+x+3 over square root of 9-x+3.

Now if I distribute this, this is what I get. The square root of 9 + x times itself is 9 + x and this is the difference of squares. Those two multiplied and those two multiplied they add up to 0 and -3 times 3 is -9. Down at the bottom, I didn't multiply this x through and there is a reason for that. Look what we got on the top here. 9 plus -9. 9 plus -9 is 0, it's gone. Look at this little beauty here, x divided by x. X divided by x is 1.

Now let's see what happened. Before, if we put 0 in here we'd be dividing by 0, it's impossible but watch what happens now when we try to divide by you put that 0 in. I'm actually just going to substitute it in and see what happens. So we've got one at the top and in the denominator we've got the square root of 9 plus, I'm going to put a 0 in for the x + 3. Square root of 9 + 0 is 3. It's 1/3 plus 3. You can do that. The answer is 1/6. We've got just a little bit more to cover in this segment. This is is one to pay particular attention to. As scary as it looks, this is actually an easy problem once you know what to do. In one of our episodes we've gone over the manual derivatives. You might want to go back and review that. That's where this definition comes from.

With a derivative, the definition of the derivative is that it's the limit of, as x approaches 0, of all of that. So the function of x plus change x minus f(x) over change in x. You wind up dividing by 0 but still produces a slope. Compare these two. That's the trick. It's just recognizing, hey wait a minute. This is just the definition with some steps substituted into it. It looks like f(x) is x³. And just to verify, look at this. F(x) plus change of x to change of x is the same variable so down here, h is the same variable as that.

This right here is the equivalent of that. Substituting x + h into the function x³. The function is x³. That's not going to be our final answer though because we're not actually doing the function, we're finding the limit of the function. The limit of the function is a derivative so all I need to do is use the shortcut for the derivative. Well, you've been doing this in your sleep, probably since the beginning or middle of first quarter. Derivative of that is 3x². That's it, that's the whole problem. It's hardly anything to do once you recognize, oh yeah, that's a derivative.

Let's do one more of these because there is another variation on this that takes one step. That's got the limit approaching 0 so it's got that part of the manual derivative. It's got that on the bottom. It looks like the function is cosine but there is no x in there right now. But if there were an x there, there'd have to be an x here as well.

This is the derivative of cosine. So let's see, our function is, cosine of x. We'll worry about the pi/2 in a minute. Since we are doing the limit of that whole mess, you know that the result has to be the derivative of cosine x. With this one don't forget, the derivative of cosine is negative of sine. We're almost done. If I replace the x with what it was replaced with in the original problem pi/2, I get f'(pi/2) equals -sine pi/2. Pi/2 is in radians of course. And the sine of pi/2, that's 1. So the result for this is -1. The negative came from the derivative.

If you ever run across a limit that you can't do and you're starting to feel sick, go to the hospital. It's actually pronounced L'Hopital's Rule and it's named after the Mathematician who invented it a couple of hundred years ago. It is really neat. The reason you didn't do it at the very beginning of the year, when you were first investigating limits is that, doing it requires derivatives and derivatives are based on limits. But it's a really cool thing. It can save you a ton of time.

What L'Hopital's Rule basically says is, if you're doing a limit of something that's a fraction, you can do the derivative of the top and the derivative of the bottom separately and the limit of that result is going to be the same as the limit of this. Don't make the mistake of using the quotient rule on it. L'Hopital's Rule doesn't use the quotient rule. So, we have the limit going to infinity.

The limit of ln to the 5th power. That's going to take chain rule, I threw that in to see if I could catch you. The derivative of a natural logarithm is 1 over whatever the logarithm's acting on. That's 1/x to the 5th but then the chain rule says we have to do the derivative of x to the 5th. So that's times 5x to the 4th. Down at the bottom the derivative of x to the 4th, we're doing it separately, is 4x³. Let's take care of a little bit of simplifying because we couldn't substitute infinity in here because we would have had infinity divided by infinity and we still do. But let's simplify a little bit, see what happens.

You've got the limit as x approaches infinity. 5x to the 4th divided by x to the 5th is 5/x and here I have 4x³, still a little more simplifying to do. So we got the limit. I'm going to be lazy and not write in the x approaching infinity. 5/x divided by 4x³ turns out to be 5/4x to the 4th. Now, we don't have infinity divided by infinity. We've got finite divided by infinite. The result, is 0.

This lesson started off with an interesting direction to the basic limits, moving on to some other more complex limits that you might see on the AP test, followed by a few of the problems that involve recognizing a derivative which you're very likely to see on the AP test.

And finally finishing up with L'Hopital's Rule which is a great shortcut when you can use it. I'd really recommend though that you download the supplementary materials from the website. Try some of the trickier limits just to get this cemented down your mind.

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