 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture I called Amazing Limits. Well, it's kind of pretentious name, but anyway, it's a couple of very important limits which which are unexpected, I would say, and there are kind of a symbiosis of different parts of mathematics, which I consider very important. Well, this lecture is part of the course called Advanced Mathematics for Teenagers and High School Students. It's presented on unizor.com. I suggest you to watch this lecture from this website because it has a very extensive explanation for each lecture, like notes, and basically it can be read as a textbook. Plus registered students can take exams, for instance, and there are some other benefits. The site is completely free, by the way. All right, so Amazing Limits. So we have two limits, which I would like to talk about today. Both are in some way amazing because personally, I didn't expect them to be in such a nice mathematically elegant fashion. But anyway, I hope you will join this kind of my amazement with these two limits, and I'm going to present it and prove, actually, that they are true. All right, the number one is and actually I did mention it before, but I will basically repeat it because it's really about limits right now. Limit of sin x divided by x when x goes to 0 is equal to 1. Well, let me just illustrate it graphically. If you have a graph of sin, and this is so this is sin of x, and this line is y equals x, which is basically angle bisector at 45 degrees. Then in the immediate neighborhood of 0, when x goes to 0, you see, they are really very much close to each other so that their ratio is equal to 1. So that's what it basically means. The design in the neighborhood of x is equal to 0 behaves like a straight line basically. Obviously, I'm talking about x being measured, it's an angle measure, right? So I'm talking about regions, and that's one of the very important actually factors why radians actually were introduced. I mean, whenever you measure angle and radians, then this line would be exactly at 45 degrees. The tangential line to the graph of the sin would be at 45 degrees, and it looks good, right? I mean, if x is measured in degrees, for instance, then you have to introduce something like a hundred and eighty factor, and that's not really nice. This is a very nice. That's why it is one of the amazing limits. All right, so how can we prove it? Well, the way how I will prove it is well, it involves geometry and trigonometry. And that's another reason why the whole thing is amazing because you see there is a trigonometry, geometry, and function analysis, limits, etc. So it's some kind of a symbiosis of different things together. So let me prove it, and basically, x can go to a zero in any fashion. In my proof, I will use the positive x, but it was a negative axis, exactly the same thing, or mixed, basically. So let me start from the definition of sine now. As you remember, trigonometric function of sine is defined on a unit circle, so this is one, and this is angle x regions. It's defined basically as the y-coordinate of the point on a unit circle, which makes this angle towards this direction of x counter clockwise. All right. Okay, so a, b, basically the length of a, b is the value of the sine of x. By definition, so there's nothing to prove here. All right, fine. So let me do the following. So this is c, and I will also put a vertical line here, perpendicular to the x-axis, and I will extend here to g. So what do I have? I have that a, b is sine of x. Fine. Now, I also connect a to c. Now, c, d, what is c, d? Well, think about it. You remember that there is also a function called tangent, and in a triangle it's opposite characters towards the adjacent characters. The adjacent is equal to one. So c, d divided by o, c, which is equal to one. So it's only c, d is basically a tangent. So this is tangent. Now, now I will compare three different areas. Areas of triangle o, a, c, sector o, a, c. Sector o, c is basically part of the circle. So it's triangle plus this little thing. So the circle is bigger than the triangle o, a, c by this particular, but by the area of this. And also, so this is a sector. And this is triangle o, c, d. Now, what can I say about areas of these three figures? So let's call this a, b, c. Now, triangle o, a, c, triangle is a, it's obviously less than sector o, a, c, because sector has this little thing. And the sector o, a, c in turn is smaller than triangle o, c, d, right? So I have this equation. Now here is an interesting consideration. I'm using my drawing basically. Whenever you are proving something using the drawing, you really have to be very careful. Because sometimes drawing can be done in one way, can be done in another way. But I hope that number one, whatever I'm doing here right now, can be extended to any other kind of drawing. Let's say for a negative x or something like this. That's number one. And number two, even if might seem to be to some mathematical purist, not a hundred percent strong, not a hundred percent rigorous, it's still not the purpose of this course to be like a hundred percent rigorous. 99 would suffice me. And I think that this presentation related to this visual kind of thing to drawing is much better from pedagogical standpoint. So you will understand that you will feel that this is kind of a nature, natural kind of thing. Because if I will start manipulating with formulas, I will probably get to the same thing, but it would not be as visible and as well understood. Let's put it this way. All right? So I will use the drawing as it is. And from this drawing it is obvious. That's maybe not exactly rigorous word. It's obvious that the area of triangle OAC less than sector OAC and less than triangle OCG. Now let's just calculate what are these three areas. Well, okay. A is equal to base is equal to one times height AB, which is equal to sine divided by two. So it's one half sine x. Now what is the B? Well, if this is an angle of x regions, now the total circle is two pi regions, right? So this sector takes x divided by two pi part of a circle. And the area of the circle is pi r square, r is equal to one. So that's my area. So it's x over two. So B is equal to x over two. Finally, C, that's area OCD. Well, it's a right triangle. The calculus OC is equal to one and the height is equal to tangent. So I multiply one by another and divided by two. So it's one half tangent x. Well, basically that's it. I don't need anymore the drawing or anything else. Drawing has served its purpose. Now I'm just using this one. One half sine x less than one half x less than one half tangent x. Now tangent x is sine x divided by cosine x, right? Now I can obviously reduce it or multiply by two. So I still have this particular equation. Now what do I do? I think the best way is, now these are all positive numbers because I'm in the first quadrant, right? So let's just invert the whole thing. If this is less than that, inverted would be greater, right? So one over sine x would be greater than one over x, greater than tangent, which I will put cosine over sine x. Now I will multiply everything by a positive sine x. So I will have this. I will have this. And I will have this in the sine. Now I will have this, this. So I have one greater than sine x over x, greater than cosine x. Now, now let's remember x goes to zero. When x goes to zero, cosine x goes to one. There is no, you remember cosine, right? Cosine again. Back to my unit circle for an angle, cosine is an x coordinate, right? So as angle diminishing to zero, my x coordinate goes to closer and closer to an entire radius, which is one. And on a graph, on a graph, graph of cosine is this, right? So at zero, obviously, it's one. So this goes to one. This is equal to one. And now let's recall the squeeze theorem, which I have proven in one of these lectures about the limits. Squeeze or the theorem about two policemen and a drunk man, when two bounding variables are going to the same number in a limit, then something which is in between has absolutely no other way but to go to the same limit. So that was a squeeze theorem or two policemen and a drunk theorem. So that's why this proves that sine x over x goes to one whenever x tends to zero. And the proof. So that's it. That's my first amazing limit. Okay, my second amazing limit is one plus x to the power of one over x. The limit as x goes to zero equals E. Now, I have introduced again in this function limit chapter of this course, what is number E? Number E, I will define the definition of number E right now, because I will be using this to prove this particular thing. I mean, obviously, if I want to prove that this goes to E, I have to specify what exactly E is. All right, so how did I define the E? Let me remind you this. If you have function y is equal to a to the power of x, I have proven that with a is equal to two, the tangential line would be at angle less than 45 degree to the x axis. And with a is equal to three, it will be much, much steeper. It will be the angle of tangential line will be greater than 45. So somewhere in between, between two and three, there is a number which can be used as a base when the line will be at 45 degrees exactly. And that's exactly how I define the number E. So let me just redraw this. So first, let me put a tangential line at 45 degrees. And now, this is how my y is equal to e to the power of x goes. And this is one. So tangential line at one is, now it's basically parallel to, it's parallel to y is equal to x, right? Now, that's how E was defined. Now, speaking about tangential line, I really have to define it better. And again, I did it when I introduced all this stuff. We can now basically express the functional definition of the tangential line. Here is how. If we are talking about tangential line at some point, let's say here, point x is equal to r. Now, tangential line can be defined as following. Let's just step, let's say forward from r. We will put something like r plus d, all right? And connect these together, these points. Now, what is the tangent of the tangential line? That's increment of the function, which is f of r plus d minus f of r. So this is increment of the function divided by increment of the argument divided by this. Now, I will do the limit as d goes to zero, which means I'm making this point closer and closer. So my chord will be closer and closer to the tangential line, which has only one common point with the curve. So now I have two points, but when these two points are closing to each other, I will have only one. So this is basically the tangent of the tangential line at point x is equal to r. Now, let's go back to my exponential function with the base e. So let's step from zero. We are talking about tangential line at point zero. So let's step forward to a point x. Now, this is the chord. Tangential line will be e to the power of x minus e to the power of zero, right? Divided by the increment of the argument. So this is increment of the function. This is increment of the argument. So tangential tangent of the chord will be this. Now, if I will do the limit of this as x goes to zero, my x point is getting closer and closer. Then my chord will be closer and closer to the tangential line. And the limit of this is actually the tangent of tangential line. And I'm saying that this is equal to one, right? Because the tangent of this line at 45 degree is one. It's parallel to this one. Increment of function is equal to increment of the argument. So this is basically a defining characteristic property of e. I actually should consider this as given. So this is given, and this is what I have to prove. Okay, this is a completely different language right now. Now, I really know what e is in this particular thing. And I know that e is a number which satisfies this. And this is given. All right, fine. Now, let me present this graphically. By the way, e to the power of zero is one, obviously, right? Because any number to the power of zero is one. So that's what I have as given. Now, again, I will approach this graphically, which again, some purists might consider not 100% rigorous. But I think it's very nicely explained actually the meaning of this relative to this. Okay, so here is my graph. Now, e to the power of x is here, right? But I'm subtracting one from it, which means I'm shifting it down. And now, it will be down to the minus one, the left part of it. And then it will be something like this. That's my e to the power of x minus one graph. And this, so this is e to the power of x minus one, this is x, y is equal to x. And not that my drawing is perfect, let me just make it slightly better. So it seems to be more like a tangential line. Okay. All right, so this is how it looks. This particular thing is the ratio between this and this. And if you remember, we just talked about sine, when the sine and x are very close to each other, this is exactly the same type of closeness. Now, let's analyze this particular thing. How can I slightly change maybe one plus x to the power of one x tends to e. Now, if something tends to something else, and I apply the same function to both of them, and I'm talking about function called logarithm, then the result will be also that the value of the function of this will be approaching the value of the same function of this. So what I'm talking about is logarithm of one plus x to the power of one x will tend to logarithm of e. Now, ln is natural logarithm. It's logarithm at base e. Logarithm can be at any base, right? So I choose the same number e as the base of this logarithm, y, for a very simple reason, because logarithm of e is one, because what is the power I have to raise e to get e? 1. So this is 1. Now, logarithm of something in the power of something else, you know there is a property of logarithms. Logarithm of a to the power of b is equal to b logarithm a. Doesn't matter what kind of a base. It can be 2, it can be 3, it can be e, it can be anything. So instead of this, I can put this. 1x should be here, outside of logarithm. So this is what I have to prove, right? Now, let's see about two things. Well, let me just put it down as a denominator. So this is what I have given, basically. And this is something which I have to prove, and they are very close to each other. I mean, they look very much like. Now, let's compare logarithm of 1 plus x and e to the power of x minus 1. These are two different functions, right? Now, let's resolve this function as x from y. It will be what? 1 plus y is equal to e to the power of x, and x is equal to natural logarithm of 1 plus e. Power, which I have to raise e to get this, right? That's what it is. And look at this, plus x, sorry, plus x. No, y. I'm talking about reverse function, y. So x in terms of y. So that's what it is. Now, what does it mean? If this function looks exactly like this, which is a result of the x resolving as y, this is inverse function. So this logarithm of 1 plus x is inverse to e to the power of x minus 1. And you remember that inverse function, you remember what inverse is, right? So to speak. Or f of g of x is equal to x. If we apply, for instance, x square and square root of x, or in this case, e to the power of x minus 1, and then logarithm of 1 plus x. If we apply these two functions in a row, we will get x again. So that's what inverse function is. And with inverse functions, if you remember, and if you don't, you can actually refer to corresponding algebra lecture, which is in the course. They have graphs symmetrical relative to bisector of the main angle on the coordinates. So what I'm saying is that the graph of this function should be symmetrical to the graph of this function e to the power of x minus 1 relative to this. And it will look like this, obviously. That's what my symmetry is, right? And it will be some kind of line where it goes to. Now, not only graphs of the functions will be symmetrical relative to this, but also tangential line to this will also be symmetrical to corresponding tangential line. So let's talk about 0.0. 0.0 in this symmetry relative to this bisector, angle bisector, reflects to itself. Now tangential line, as we know for this function, is the bisector, right? Now bisector reflected to bisector again will reflect into itself. So what happens is that this point in this reflection stays and the tangential line also stays. And this tangential line would be tangential in exactly the same way to this one. So this is basically the basis, so to speak, for the proof which I'm suggesting right now. Since my tangential mechanism remains in place, the same line which is dividing my angle in two halves to equal halves is tangent to this and tangent to that. Then there is absolutely no difference in the result of this limit. So basically I'm considering that I have, from this as given, I can say that in exactly the same fashion this is given as well. Where I replace my e to the power of x minus one with inverse function and and I did it because they are completely symmetrical and the tangential line to one of them is reflecting to a corresponding tangential line into another, but they are the same since they are coinciding with the axis of symmetry. So now since this is proven, how can I go back to this? Well easily. First of all, if I know that one x logarithm of one plus x tends to one, now I can put back logarithm inside to the power of one x. That's the same thing. It also goes to one. Now if this goes to one, then if I will use exponent e to the power, I will get e to the power of left part and e to the power of the limit. e to the power of logarithm which is natural logarithm with a base e would be one plus x to the power of one x, one x and the limit will be e to the power of one which is e. So I have this. Again, if we will excuse my usage of the graph and inverse function etc as maybe only 99% rigorous, then it's pretty good explanation of why this thing is really happening and I think it's much better kind of approach which gives you really very visual representation of why this is actually true. So I think this is actually kind of a better approach to give you the feeling of why this thing is happening. Not only this thing is happening which we can probably prove by some chain of formulas one after another, but this actually shows you what's behind it because most likely the real mathematician first find this type of laws using some intuitive methodology and whatever I was showing is basically just an expansion of intuitive understanding of this and only then if they really want to they they can approach it in a very very strict and rigorous fashion with formulas etc etc. So I would like to stop here. I'm not going any to any more details about these type of things. The only little consequence of this is which I probably mentioned in one of my lectures before if you will take 1 plus 1 over n to the power of n where n is just a natural number obviously it goes to e as well because if n goes to infinity 1 over n goes to 0 so if x is equal to 1 over n then n is equal to 1 over x right so it goes exactly into this formula so limit as n goes to infinity is the same as this where x is going to 0 with this substitution and this is actually a very interesting thing I mean you probably have heard about this at 1 plus 1 nth to the nth power goes to some strange number e and by the way this can be another definition of e so we can actually use this limit or this limit whatever as a definition of e and then derive the other thing like e to the power of x minus 1 divided by x goes to 1 so it's basically all these equivalent definitions of e can be replaced one with another using one as a definition and another is supposed to be proven or the other way around okay that's it for today thank you very much and I suggest you to read the notes for this particular lecture just to refresh your memory and and maybe better understand it other than that you got it thanks and good luck