 Hi, I'm Zor. Welcome to a new Zor education. I'd like to solve a few problems related to limits. And in this particular lecture, I will consider limits when the argument of the function goes to some finite real number. Now, as usually my kind of preamble is that this lecture is part of the course of advanced mathematics for teenagers and high school students. It's presented on unizor.com. I do suggest you to watch this lecture from this website because it has lots of notes and exams for registered students. This website is free. You can take as many times exams as you want. And another important thing is I do suggest you to go to the website to notes for this lecture where all these problems are presented and try to solve them yourselves. That's probably the most important part. You do something yourself and then, only then, you listen to this lecture and check if your answers are more or less the same as whatever I will come up with. Alright, so let's do it. Now, what are we considering right now? We're considering the fact that the function f-attacks should go to some limit l as its argument goes to some real number a. Using the epsilon-delta language, which we were considering before a few times, what it means is that for any positive epsilon, however small, we can always find such delta neighborhood of a such that if my x is within this delta neighborhood of a, so the absolute value of the difference is less than delta, from this immediately follows that f-attacks would be in epsilon neighborhood of its limit l. That's what basically this means. This is just words and this is the exact mathematical definition of the limit. So, in this particular lecture, I will just exemplify how different limits can be calculated best in whatever the properties of the limits we know. We know a lot of properties. The first few examples are related to limits of a very simple nature. If you have a continuous function and then your argument x equals a and this is f-attacks equals l, for instance, this is l. So, if your argument goes to a, then the function basically goes closer and closer to this value l and the function f-attacks goes to its value at point a for continuous function. So, if your function is continuous, then instead of basically calculating what can that limit be, you can just go and use the value where x actually is approaching as an argument to a function and that would be the limit. And it's based on the continuous property, continuity of the function f-attacks. And most of the functions which are considered in school are continuous except certain cases when they're not obviously, but most of them are and I will just offer a couple of examples of these type of functions. So, the first example is a polynomial 3x3 minus 2x squared plus x minus 1 as x tends to 1. So, the polynomial functions are always continuous. So, in this particular case, you can just substitute the limit value of the argument into the function and you will get 3 minus 2 plus 1 minus 1 which is equal to 1. So, 1 will be the limit of this function. This function actually is going to the value of 1 as its argument goes to 1. And if you will draw some kind of a graph of this particular function, you will see that graphically it's actually the same thing. At x is equal to 0, it's minus 1. At x is equal to 1, it's 1. So, it's something like probably it's a cubical parabola they're saying and it looks like this. Since this is a positive, it goes to positive plus infinity if x goes to positive infinity. So, that's approximately the graph of this function. And it's obvious that as x approaching 1, 1 is supposed to be here. 1. As x approaching 1, the value of the function approaches to 1 as well. So, basically that's the simplest way, simplest kind of limits when you don't really have to think about how to calculate it. Just substitute the value of the argument, the limit value of the argument into the function and you will get the limit of the function. Now, another example of the same kind when you can just substitute into the function instead of argument its limit value is 2 to the power of x as x goes to 3. Now, same thing, 2 to the power of x looks something like this. So, x is equal to 3, function is equal to 8, 2 to the power of 3 and function is all the exponential functions are continuous. So, you can just use the limit value of argument substituted in function 2 to the third degree would be 8. So, this function goes to 8. That's simple too. What else do I have? Okay, trigonometric functions, sine of x. In this case, x goes to p over 2. Now, again, sine is something like this, continuous function. This is 0, this is p over 2, this is 1 and obviously you can just substitute p over 2 into the sine as a function and you will get the value of 1. So, this function would go to 1 as x approaching to p over 2. There are no problems with continuity in this particular case. So, the limit of function is equal to function of limit of this argument. And the last one which I wanted to consider is logarithmic function also continuous. So, decimal logarithm of 10x as x goes to 100. Logarithm looks like this. This is 1, this is 0. Now, this is decimal logarithm with a base 10 and it's a continuous function so you can just substitute and you will see that the limit is logarithm of 10 times 100 which is 1000. Decimal logarithm of 1000 is 3 because 10 to the third degree is 1000, right? 10 to the power of 3 is 1000. So, logarithm is the exponent which should be used with the base 10 to get 1000. So, this is 100 so this would be 3. Obviously not to scale but anyway. Okay, so these are all examples of different continuous functions polynomial, exponential, trigonometric and logarithmic. So, if you have just these simple functions or a simple combination of it then basically you are in good shape. Now, let's consider cases when it's not so obvious and my first case is the ratio of polynomials as x goes to 2. Now, what's the problem in this case? You cannot substitute 2 into this particular function because it will be 0 on the top and 0 on the bottom, right? So, basically it's indeterminate as they call it. So, what would be the graph of this function? Well, let's just make a couple of simplification before. So, you see, in cases like this you really have to think about how to simplify your function in such a way that the result would be more palatable and more understandable about how to take the limit of it, right? So, what is the simplification which can be done in this particular case? Well, it's obvious like this. x square minus 4 is equal to x minus 2 times x plus 2, right? It's the difference between two squares. a square minus b square is equal to a minus b times a plus b. The a is x and b is 2. So, you know that and that's very interesting because now you can say that x square minus 4 divided by x minus 2 is equal to x minus 2 and x plus 2, right? And you have x minus 2 in the denominator. So, that will be x x plus 2 for all x not equal to 2, right? So, this is equation, equality actually for every x which is not equal to 2. And 2 is exactly the limit point, right? Now, if 2 is a limit point, that's okay. We are not equal to 2. We are approaching 2, but we are not equal to 2. So, it means we are actually approaching to number 2 using this function. And this function is just a straight line, right? This is x, so this is x plus 2. And 2 is here. So, if x is 2, it goes to 4, obviously, right? So, this limit is equal to 4. 2 plus 2, that's 4. And again, I have just substituted instead of x square minus 4, it's different representation and I was able to cancel x minus 2 for every x which is not equal to 2. And that's quite alright because 2 is not part of my limit. It's a limit point, which means we can do this. For every limit point, wherever x is not equal to 2, we can calculate it. So, that's the first example where you have some kind of indeterminate limit and you have to do something, some kind of a transformation simplification, whatever you call it, to get to something which really is much simpler. My next example is this, 1 minus cosine of x divided by x square. Now, I would like to remind you a couple of so-called amazing limits. So, if you remember, sine x divided by x goes to 1 as x goes to 0. And another example, which is actually a definition of number e, e to the power of x minus 1 divided by x also goes to 1 as x goes to 0, if you remember. So, these are couple of amazing limits which I have already explained in some other lecture. And I will use them, obviously. Now, this is cosine, this is not sine, right? But, however, again, let's just think about it. You see, if you have a sine over x, remember, sine looks like this and x is equal to this. So, they are very much close to each other and that's why the ratio goes to 1. In case of cosine, that's not exactly true. Let's put graph of cosine. Now, this is my cosine, around 0.0, right? Now, we need 1 minus cosine. So, if this is 1, so 1 minus cosine would be very close to this. It would be something like this. Now, what does it look like? Well, it looks like a parabola, right? It's not like a straight line. And what it means, it means that there is a chance that they are actually close to each other. The ratio one against another might also be close to some constant. And to basically establish this, now, this is my kind of analytical research. You see, I do see the shape of the 1 minus cosine and I remember the shape of the x square, a parabola, and I think that they might actually be close to each other. All right, so let's just see if that's true. And that's exactly why this is x square and not plane x, because plane x is definitely not good in this particular case. Plane x would be looking like this. This would be my 1 minus cosine and x would be like this. Obviously, I cannot expect the ratio to be anything but 0. In this particular case, I have the hope. So how can I come up with something which is basically allowing me to do this type of thing? Well, let's just remember that cosine of x, which is cosine of x over 2 plus x over 2, is equal to cosine square of x over 2 minus sine square of x over 2. Or if I replace cosine with minus sine, it will be minus sine and another sine will be 2 sine of square of x over 2, right? Now, 1 minus cosine of x would be 1 minus this, which is 2 sine square x over 2, right? Now, we have a very interesting situation. We have this equal to 2 sine square x over 2 divided by x square. Alright, this is closer to this one. Well, I need something square divided by something the same. I have to have the same argument, right? So I will do it this way. 2 sine square x over 2 divided by x over 2 square. Now, I have the same arguments, right? As here, x and x, right? And since I have the same argument, now what did I do? Now, this is x square divided by 4. 4 goes there, so I have to multiply this by 4, right? That would be correct. Okay, now I can represent it differently. Now, I can represent it as 1 half sine of x over 2 divided by x over 2 square, right? That's the same thing. And I know that as x goes to 0, x over 2 obviously goes to 0 as well. The whole thing goes to 1. Square would be 1, so the whole thing would be to 1 half. This 1 half. So that's how I calculated this particular argument, all right? Move on. Let's move on. Okay, what's next? Next is 5 to the power of x minus 1 divided by x as x goes to 0. Now, obviously I have to somehow use this one, right? So how can I get from here to there? Well, here is how. 5 is equal to e to the power of logarithm, natural logarithm of 5. Natural logarithm is logarithm with a base e, right? So that's basically the definition of the logarithm. e to the power of logarithm 5 is equal to 5. So now I can have 5 to the power of x minus 1 divided by x is equal to e to the power of logarithm 5 to the power of x, right? Minus 1 divided by x, which is equal to... Whenever I'm having two different exponents, one after another, the exponents are multiplied, right? So it's x times logarithm 5 minus 1 divided by x. Now, if x goes to 0, x times logarithm 5 also goes to 0. So my easy way out is times logarithm 5 here and times logarithm 5 there. Now this goes to 0 and this goes to 0, so I can use this. It doesn't matter whether it's x or x multiplied by some value. It still goes to 0, right? So this goes to 1 and the whole result is this. So whenever I have this, it goes to natural logarithm of 5. Next, what's my next? Okay, next is sin x divided by e to the power of x minus 1. Well, obviously you just have to multiply it by x and divide it by x. Well, if x goes to 0, of course. And just group it differently and it would be sin x divided by x divided by e to the power of x minus 1 divided by x. This goes to 1, this goes to 1 and that's why the whole thing goes to 1. I cannot put 0 in the very beginning as a limit point because sin of x would be sin of 0 would be 0 divided by e to the power of 0 which is 1 minus 1 also 0. So it's again indeterminate. But after this I have everything determined because now I have basically the division of two functions each of which I know where they're going, where exactly they're tending to. Okay, and more. Next example is 1 over x square minus 3x plus 2 minus 1 over x square plus 5x minus 6. Now what's wrong with this? Well, if x goes to where? To 1. x goes to 1. So what's the problem with x going to 1? Well, this is if x goes to 1, 1 minus 3 plus 2, that's 0. So it's 1 over 0, it's kind of infinity, right? Now this 1 plus 5 minus 6 also 0. So it's another infinity. Now we have infinity minus infinity. So we have two different functions, each of them infinitely growing and the question is how their difference will behave. We have absolutely no idea. So to do this, the best way is just basically get to the common denominator and see what happens. So what is the first one? The first one can be expressed as x minus 1 times x minus 2, right? And the second one would be x minus 1 x plus 6. Am I right? Right. Now the common denominator is x minus 1 times x minus 2 times x plus 6, right? So I have to multiply this x plus 6 and this I should multiply x minus 2 and I have common denominator. Now what happens? Well now we don't have minus. Now we don't really have the problem because on the top I have 6 minus minus 2, I have 8 and on the bottom I have only 1 infinity because if x goes to 1, now this is going to minus 1, this goes to 7 so these are constants basically and only this one goes to 0 which means the whole thing basically does not have any normal limit, right? So basically the answer to this is the function has no limit because we really don't talk something like function has a limit of infinity plus we don't really know which infinity, positive infinity or negative infinity. If we are talking about positive or negative then we can say that the function is infinitely growing or infinitely diminishing but in this case we don't even know the sign of this because if x goes to 1 from the right, being greater than 1, this would be positive. If x goes to 1 from the left, from the 0, it would be negative so the infinity would be negative which means it all depends on how x is going to the 1 which means that there is no limit basically, that's it. So that's the case when you can say that there is no limit. Unless somebody will tell you, okay, x is going to 1 something like this plus which means it goes to 1 from the right or x goes to 1 minus from the negative side then you can say that in one case function would be unlimited growing or unlimited diminishing but if this is not said, if it's just like this, then the function just had no limit and that's fine. Next, natural logarithm of sin of x minus natural logarithm of x. Well, this is actually easy because we obviously know that the difference of logarithm is a logarithm of ratio, right? So this is equal to logarithm of sin x over x. I didn't specify x goes to 0 in this case. Now, if x goes to 0 then you remember the compounding function. So this obviously goes to 1 so the whole thing goes to logarithm of 1 which is 0. And the last one is another little example. Square root of x plus 1 over x minus 4 minus square root of x plus 16 over 2x minus 8 as x goes to 4. Now, again we have this situation when you have at the limit point this is equal to 0 and this is equal to 0 which means we have some kind of indeterminate one infinity minus another infinity and we have no idea about what's the difference will be. So somehow we have to manipulate this particular expression to come up with more palatable form of this. So obviously x minus 4 and this is 2 times x minus 4 so that should become a denominator. So let's start with this. So it would be what? It should be 2 times x minus 4. That's the common denominator. This would be 2 square root of x plus 1 minus square root of x plus 16, right? I multiply this by 2 so I have the common denominator. Now what can we do with this one? Well, a typical example of how to deal with this situation is this and what we usually do we use this formula. Now this I consider to be this. So let's multiply it by plus and then we will have a difference of the squares. So let's multiply by 2 square root of x plus 1 plus square root of x plus 16 and divide by the same thing. 2 square root of x plus 1 plus square root of x plus 16. Now what happens now? Now this is a minus b, a plus b. So it's equal to a square minus b square which is 4 times x plus 1 minus x plus 16 divided by 2 x minus 4 2 square root of x plus 1 plus square root of x plus 16. Now what is this? 4x minus x is 3x, 4 minus 16 minus 12. So this is 3x minus 12. Well, let's factor out 3 and we have 3 times x minus 4, right? And this is exactly what I was going to do. So x goes to 4, right? Which means it's not really equal to 4, it's just approaching 4. And that's why we can do this safely. And what's left? Well, what's left are basically functions which do not have any indeterminateness among them. So we can just substitute 4 and we will get what? 3 over 2 times 2, so that's 4, square root of 5 plus 2, square root of 4 plus 16, 20. Well, square root of 20 is 4 times 5, square root of 4 is 2, so I can say 4 square root of 5. And then I can obviously combine them together and I will have 8. And that's the answer. So that's the limit. As you see, there are different techniques which I was using. What's very important is for you to basically be familiar with these techniques, to be able to know them, to be able to apply them whenever you want and to be able to see actually which technique you should use. And for this, I do suggest you to go back to Unizord.com, look at all these problems and go ahead and just try to solve them yourselves. Well, that's basically it. Thank you very much and good luck.