 And now we come to something so interesting, the indeterminate form. Now, I warn you, you must have done a bit of calculus before just taking first derivatives, sometimes repeating that first derivative, in order to do these. But usually it's not too difficult. If you've had some high school calculus, you can do simple first derivatives. You should be okay. Look at these two expressions, t plus 1 over e to the power t and t minus 2 over t cubed minus 8. And if I were just to do direct substitution there on the left hand side, I'm going to get infinity over infinity. On the right hand side, I'm going to get 0 over 0. These are not values. These are not numbers. You can't leave it as such. Now, we've seen this before. What we did is we divided the numerator and denominator by the largest power variable in the denominator. But if you look on the left hand side, we have an exponential there. A normal polynomial. In other words, we're hitting a bit of difficulty. For now, all I want you to do is imprint those two things in your mind. Infinity over infinity, that can be positive or negative. And 0 over 0. Those are solutions that we call indeterminate. These limits are in indeterminate form. And there's something we can do about that. And this is what we do. The L'Hopital's rule. And that says if you have a fraction, one polynomial divided by the other. Now, f of t divided by g of t, with g of t being defined. And both of them in the limit is 0 or infinity. Remember, infinity positive or negative. What you do according to this rule is to take the first derivative separately of the f of t, separately of the g of t. We're not going to make use of the quotient rule here. This is not the first derivative of the whole quotient f of t divided by g of t. This is something I would almost want to say un-mathematical. We're just going to take very separately the first derivative of the numerator and the first derivative of the denominator. That is quite different from the quotient rule. Do not get the two things confused. Now, L'Hopital's rule actually goes much further. You actually have to do tailor polynomial, tailor expansions, et cetera, et cetera. But we're not going to deal with that kind of thing here in limits. So let's have this. The limit is t goes to infinity of t divided by t plus 2 squared. If I were just to do the substitution, I'm going to end up with infinity of infinity. That's indeterminate form. So what do I do? What is the first derivative of the variable t? Well, that's just 1. The first derivative of t plus 2 squared is 2 times t plus 2. I show that on the right-hand side. And now, if I do that limit, I get 1 over infinity. We've seen that before so many times. That equals 0 exactly as the graph described before. So we do have a solution for this limit. Let's look at t divided by e to the power t. You can see the graph there. It's going to go to 0. It's infinity of infinity in determinate form. We take the first derivative of the numerator, which is 1. We take the first derivative of the denominator, e to the power t. It's first derivative is e to the power t. Again, I end up with 1 over infinity, which is 0 beautiful. What about t squared? What I'm going to show you here is sometimes you have to apply L'Hopital's rule twice because the first derivative of t squared is 2t. Again, over e to the power t, if I do the limit now, as t approaches infinity, I'm stuck with infinity of infinity again. Again, in determinate form, what do I do? I apply L'Hopital's rule again. And now I end up with 2 over e to the power t. Again, the limit of that is 0. And I show you there, I take the constant 2 outside of the limit. It doesn't matter. It ends up being 0. So those are all easy to do, but these are the ones that you're going to be faced with. Now this is not truly in determinate form. If you get a solution of one of these five, you have to do some jumping around. You're going to have to do a lot of algebra. I warn you, some of these can be very tricky, very difficult. But you're going to have to do something to your expression to change it so that you get either infinity over infinity or 0 over 0. Those are the two in determinate forms. These are not. You cannot apply L'Hopital's rule when you get this kind of solution. And what you're going to get is one of these five. Infinity minus infinity, 0 times infinity, 0 to the power 0, 1 to the power infinity, infinity to the power 0. Those, if you get those kind of results, you're in trouble and you'll have to get out of it. Let's have a look at the first one, infinity minus infinity. There we have our expression, 1 over t squared minus 1 over t to the power 4. You put in the value as t approaches 0, you're going to get infinity minus infinity there. Now, you're in trouble and you have to get out of trouble. Sometimes it's going to be easy to get out of trouble. Sometimes it's going to be very difficult. And I'll show you how difficult it can get. Now, one of the first things you can do to get out of trouble is to use common sense. First of all, if you were to do a bit of simple algebra, just get a common denominator there. This is what you have. Now think about it. Without applying anything, just think about things a little bit. What is going to happen to the numerator? Just imagine in your mind what is going to happen. t squared minus 1, well, the t squared is going to go to 0, so your numerator is going to tend to be negative 1 as t approaches 0. Your denominator is going to become a tiny, tiny, tiny value is going to approach 0, and any constant divided by a minutely small number is infinity. But we've got the negative 1 in the numerator, so this really is going to be negative infinity. Okay, we could use common sense to work this one out. Let's look at something way more difficult, and you see there I've put down the word difficult even. Again, with infinity over infinity, we have 1 over the natural log of t plus 1 minus 1 over t. I plug 0 into both of them. I'm actually going to get infinity minus infinity. That's not going to work for me. I cannot apply L'Hopital's rule because infinity minus infinity is not true in determinant form. It's a special kind of indeterminate form. Let's do some basic algebra. We're getting a common denominator. That's becoming a theme, isn't it? So there we are in line 2. I've got my common denominator, and I see if I now would plug in t approaching 0, I get 0 over 0, lo and behold, I can now use L'Hopital's rule. I can take the first derivative of the numerator, I can take the first derivative of the denominator, pause the video, do it on your own, see if you get the same thing, and now I'm left with that. What do I get is 0 over 0, still not helping me out much. Now, one way we saw was to reapply L'Hopital's rule, and this instance is not going to work for you. You can go down that avenue, you're going to find a dead end. You have to look for something else. And look what we did at the top on the right-hand side there. Something very surprising. Very surprising. We multiplied both the numerator and denominator by t plus 1 over t plus 1. This is not something that can be taught. This is not a rule. This is something that you will have to look at. You will have to look at it and decide what to do, what will make that expression, or what will change that expression to make it into something that is again in determinate form that will help you out of your problem. You have to explore all sorts of avenues. Many of them will be dead ends. One of them won't be a dead end. In this instance, this multiplying by t plus 1 over t plus 1 is not a dead end. So we end up with the expression that you see there. If we take the limit as t approaches 0 of that, we are going to get 0 over 0 again, true in determinate form. Now we can apply L'Opital's rule. And now it's going to be very simple. We're going to get 1 over 1 plus natural log of t plus 1 plus 1. That's 1 over 2 1 half end of problem. Now you're not just going to look at this problem and solve it within a couple of seconds. You are going to go down many, many dead ends. Stick to it though. Somewhere along the line, something's going to help. So little rules of thumb. Start by getting a comment denominator. Get it in determinate form. Apply L'Opital's rule. Even apply the second time. If you run into a dead end, or it makes things even worse, applying L'Opital's rule, back up, take another fork in the road, try something else. The other thing to try is to multiply your expression essentially by 1 because t plus 1 over t plus 1 is just 1. So you're multiplying by 1. You're not changing anything. That usually gets you out of jail in some form or fashion. Let's look at this. Infinity times 0. This is an easy one. We see the natural log of t times 1 over t. If I were to just plug in infinity, I'm going to get infinity times 0. That's not true in determinate form, but I can just do a simple rewrite. Natural log of t times 1 over t is actually just a natural log of t over t. If I were to plug in infinity now, I'm getting infinity over infinity. That, I can apply L'Opital's rule. I can get the first derivative of the natural log of t and the first derivative of t. So I'll just end up with 1 over t. So that is 1 over t and the limit as that approaches infinity of t, that's just 0. So that's an easy one. Always just look out for rewriting things in a slightly different way. So another little tool in your basket there. Okay, now things are going to get very serious. This is only for someone who needs to do hardcore, difficult problems. Look at this. And that's basically the last of the tools that I'm going to put in your pocket. The limit as t approaches 1 to the power 1 over t minus 1. I'm telling you now, very difficult. There's no way you're getting out of jail by rearranging that in any form, by multiplying it out by anything, any algebraic way of tackling that problem is basically not going to work for you. Now, what do we do? We're going to introduce a new variable. So that's a whole new method you see there in red. Y equals, I'm going to take my expression t to the power 1 over t minus 1 to equate that to a new variable. I'm going to call that variable y. What I'll now do is take the natural log of both sides. So the natural log of y is going to be the natural log of t to the power 1 over t minus 1. By the laws of logarithms, I can bring that power in front of the natural logarithm sign there. So I'm going to get 1 over t minus 1 times the natural log of t. Now let's take the limit. Not the limit of y, which is my original expression, but the limit of the natural log of y. If I were to do that, that'll be the limit as t approaches 1 from the positive side of 1 over t minus 1 times the natural log of t. Now we say t from the positive side because remember the natural logarithm is only defined for values larger than 0. So we're viewing things from the positive side here. So if I were to do that, if I were to just plug in... if I were just to simplify things, I'd put natural log of t over t minus 1 and I plug in t approaching 1. I'm going to get 0 over 0. That's indeterminate form. And I can now simply take the derivative of the numerator and the denominator and that leaves me with 1 over t. If I were to plug in t approaching 1, I'm going to get 1 over 1, which is 1. So the limit as t approaches 1 from the positive side of the natural log of y equals 1. But I was never interested in the limit of the natural log of y. I was interested in the limit just of y, y being my expression t to the power 1 over t minus 1. I do remember something from mathematics though. Any variable like y, I can rewrite as e to the power natural log of y. Those two things are exactly the same. So if I were to take the limit as t approaches 1 of y, that would be the same as taking the limit as t approaches 1 of e to the power natural log of y. And I remember my limit laws now. Because if I take the limit of a constant to the power some variable, it would be the same as taking that constant and then taking the limit of the power. You'll see there it now becomes e to the power the limit as t approaches 1 from the positive side of the natural log of y. Lo and behold, look a few steps back. We've just seen that the solution to that is 1. So I am left with e to the power 1, which is e. So a lot to take in a couple of seconds. Watch this video again and again. This is absolutely beautiful. There's beautiful mathematics happening on this slide. It's actually such a cunning way of getting out of trouble. Once you've understood this, once you've mastered the use of this, it is a beauty. Let's have a look at another example where I'm going to do this. Now we're going to get 0 to the power 0. It was one of the five ones, which is not too indeterminate form and we cannot apply Lopital's rule. So it's the sign of t, all of it to the power t. If I were to plug in t approaching 0 from the positive side, I'm going to get 0 to the power 0. It's not too indeterminate form. I have to do something else. You can't do anything algebraically there. What you will have to do is introduce a new variable. The last little trick in your pocket. y equals the sign of t to the power t. I take the natural log of both sides, which allows me to get rid of the power because I can bring the power in front of the natural log sign. It's one of the laws of logarithms. I now take the limit as t approaches 0 from the positive side of the natural log of y. Not y, the natural log of y, which is the limit as t approaches 0 from the positive side of t times the natural log of the sign of t. Now, look, these two brown circles. It's just an algebraic rewrite. I'm taking that t and I'm putting it in the denominator, but to do that it's 1 over t. If I do that and I plug in t approaching 0 from the positive side, I'm going to get negative infinity in the numerator, infinity in the denominator. I said you can get plus or minus infinity of infinity. That's still L'Hopital's rule. Look, though, just at the graph at the bottom there. As t approaches 0 from the right-hand side, we see the graph does approach 1, so we better get to a solution of 1. Of course, with this kind of slide, you can see the bottom right-hand side. The solution is actually 1. We do get to it. Now, top right, in indeterminate form, we can take the derivative of the numerator and the derivative of the denominator. These are slightly more difficult derivatives. You'll have to know something about those first derivatives. We end up with that now. What now, though? I ask you, with tears in my eyes, what now? I show you now what to do. You're going to go through so many blind alleys, dead ends, until you figure this one out. If I were to multiply both the numerator and denominator by t squared times the sine of t, you can well imagine this is not a hard and fast rule. This is not something you can be taught. This is something that you do. You stay at that expression, cosine of t over the sine of t over negative 1 over t squared until you figure out something that will lead you down to the gates that open to the solution as opposed to a dead end. If I were to multiply both my numerator and denominator by t squared times the sine of t, I'm going to end up with a negative t squared cosine of t over the sine of t. If I take the limit of that to approach a 0 from the positive side, I'm going to get a proper 0 over 0. In this indeterminate form, I can apply L'Hopital's rule. Once I do that, the first derivative of the numerator ends up being negative 2 times t times the cosine of t plus t squared times the sine of t. The denominator ends up being the cosine of t. If I now were to plug in t approaching 0, I get 0 over 1, which is 0. A proper solution. Remembering this is the limit of the natural log of y, which is never what I was interested in. I was interested in the limit of y. I remember mathematically I can write y as e to the power natural log of y. I take the limit now of y on the left-hand side, which means I've got to take the limit of e to the power natural log of y. My laws of limits allow me to take that limit up to the variable itself, leaving a constant alone. I've just done it. I know that the limit as t approaches 0 from the positive side of the natural log of y is 0. Look, 2 lines up. I've done that. I can plug in 0 there. It's e to the power 0, which is one which my graph shows on the left-hand side. This is a beautiful, beautiful problem. Don't you think? Let's go into the last one. We have the limit as t approaches infinity of t plus 1 to the power 2 over t. Plug in infinity. You're going to get infinity to the power 0. It's one of the five difficult ones. You have to go down a few blind alleys until you get out of trouble. What did we do before? If we can't manipulate that algebraically, we introduce a new variable, y. We take the natural log of both sides. We take now the limit of the natural log of y as opposed to y. You see I've replaced y there with what I have, t plus 1 over 2 over t. I rewrite that because I can now bring the 2 over t in front of the natural log sign. We see it there. And lo and behold, if I were to replace t now with infinity, I get 0 times infinity. That's another one of those that I can't really do anything about. I rewrite it though. See the two brown circles? Just an algebraic rewrite. It's 2 times the natural log of t plus 1 over t. If I were to plug in t equals infinity now, I get infinity over infinity. In determinate form, proper in determinate form, apply L'Hopital's rule. And what do I end up with? 2 over t plus 1. If I take the limit as t equals 0 of that, I get 0. Once again remembering that was the limit of the natural log of y, not the limit of y itself. Same as before, y equals e to the power natural log of y. I take the limit of y. I can take the limit up into the power because e is a constant. I've just checked that the limit is t approaches 0 of the natural log of y is 0. I can replace that with 0. It's e to the power 0. And we end up with 1 as we can see the graph on the right-hand bottom. Now these are wonderful problems. Sometimes you're going to sit with them for so long, but it's a challenge. You are an investigator, a private investigator. You are Sherlock Holmes, whoever you want to be. You've got this problem that you've got to solve. You're going to sit with it. You're going to fight with it. But you've got some tools in your bag now. Try all of them. The most difficult ones are the ones that you have to decide to multiply both the numerator and denominator by something that will lead you to a path that's not blocked. Good luck with those. Enjoy them. There are fantastic problems to try and solve.