 Hi, I'm Zor. Welcome to Unisor Education. Today I will offer you a couple of problems, solutions actually to these problems. These are very, very simple things. It's just the beginning of the limits and the problems actually with solutions are on the website Unisor.com so you can read it not only listening to me but just read it from the website itself. So again, these are very simple problems. However, their importance is in approach basically because the question is how do you prove that some sequence has certain limit. So there are certain approaches to this and I would like to illustrate these approaches using these very, very simple problems. So I have five problems here so I will do it once, one at a time. The problem number one, you have a sequence of these actions. Now question is does it have a limit and if it does what is this limit? Well, obviously everybody understands that if denominator goes to infinity basically as an increasing then the fraction which has a fixed numerator should go down to zero and it does. So let me just show, let me prove actually that zero is a limit of this sequence not just intuitively but mathematically according to whatever logic and definitions we have. So let's recall what is a definition of a limit. If you have a sequence, elements are a nth then the limit of this sequence as n increasing to infinity is some kind of a number l which means that the distance between the limit and elements would be less than any d if n is greater than some capital N which we found depending on d. So the smaller the d actually is the larger probably the m should be but what's important is that once this particular sequence gets into the distance of less than d from its limit then it stays this way. So for all n lowercase n greater than capital N which we have found this particular inequality will hold. Now if I'm claiming that zero is a limit of that thing then for any d whatever the d I find I should find such a number capital N that after the order number of our sequence elements is greater than this capital N this particular inequality would hold. So l is equal to zero now 1n I should find n when this is actually held. Well that's very easy zero minus 1n absolute value is equal to 1n so 1n should be less than d. I can invert inequality and that means that I have to change the sign from less than to greater and that would be 1 over d. So it looks like for any n greater than 1 over d our inequality would hold because these are completely invariant transformations from this follows this which means that I can choose n equals to next integer after 1 over d. But 1 over d is some kind of a fraction probably whatever it is depending on d and n is integer so the next integer after that as long as it's greater than 1 over d. So it's the next integer which is greater than 1 over d would actually fit our definition because for any lowercase n which is greater than this one this would be satisfied and that's why this would be satisfied which means zero is a limit. We found for any d we have found the number n after which all sequence elements will be within the d distance from its limit. That's the problem number one. Now what's the kind of lesson which I would like you to learn from this particular proof? Go from the definition. That's very important to start what is the definition of limit in this particular case. Well actually it's very important in any kind of discussion. If two people have different opinion first of all they should discuss the definition of what they're talking about. If this definition is different if one is talking about apples and another is talking about oranges and one is saying that this is hard and this is soft then obviously they will not be able to agree about anything so they have to talk about the definition first. So we started from the definition of the limit and then we proved that whatever we have proposed as a limit actually is a limit indeed. Now similar problem 1 over 2, 2 over 3, etc. n over n plus 1, etc. So that's the sequence I'm talking about. Now I state that the limit of this as n goes to infinity is equal to 1. Now intuitively it's obvious why because n over n plus 1 is equal to 1 minus 1 over n plus 1. That's obvious, right? n plus 1 times 1 so it's n plus 1 minus 1 over n plus 1 so it's n over n plus 1. Now what it means is that the limit of n over n plus 1 is equal to limit of n minus 1 over n plus 1 and 1 minus 1 over n plus 1. Now we have learned from the previous lectures that limit of sum of two sequences is equal to sum of their limits if these limits exist, if all three limits exist. Now in this particular case 1 is not really a sequence but a constant but you can always consider any constant as a sequence of these constants repeating itself so it's 1, 1, 1, 1. And this is another sequence which we have already examined in the previous problem. It's a sequence of inverse integer numbers, natural numbers. So I can safely limit of 1 minus limit of 1 over n plus 1. Now this is obviously 1, this is 0 as we have proven in the previous theorem and basically that's enough to, as a proof that the whole limit is equal to 1. Now a different proof obviously can be suggested where I explicitly go from the definition so let's talk again about definition. So what I have to find out is that for any g greater than 0 that's the distance from the limit. I have to examine absolute value of 1 minus, if I will find such a capital N that this would be true if lowercase n is greater than capital N then that would prove that 1 is a limit. So we have chosen d, now we have find such a capital N that this particular inequality is true for all lowercase n greater than capital N. Well let's again do some kind of transformation, invariant transformation of this inequality which is basically solid for N. Now if I will convert this it would be n plus 1 minus n over n plus 1 absolute value which is equal to 1 over n plus 1 and it should be less than d right? Well obviously again we inverse the sign of inequality and we come to conclusion of this. So if I will choose capital N which is equal to next integer after 1 over d minus 1 then for all lowercase n greater than this capital N this would be satisfied because all these transformations are invariant. So it's exactly the same basically theorem just proven differently. First I proved using the limit of the difference between two sequences and this is a direct proof based on definition of what actual limit is. So that's problem number 2. Next, do all sequences have limits? Well here is an example of a sequence which does not have a limit and we can prove it. Minus 1, 1, minus 1, 1, etc. Which is basically minus 1 to the nth degree. That's where it is. So the sign is changed over time. Now does this sequence have a limit? Well obviously everybody understands that no matter what kind of number we choose as a supposed limit we will always find members of this particular sequence further than that number by whatever by any distance actually. Relatively small distance obviously. Let's choose a distance of let's say 1. So and then take any number number 8 as a proposed limit. Now if d is equal to 1, question is do we have some number n such that this would be less than d or d equals 1 in this case for all n greater than n? Well let's think about it. a can be either positive or negative right? Now if a is positive then we do have numbers minus 1 occurring anywhere in the sequence up to infinity. So no matter what number we choose we will always find numbers minus 1 beyond that boundary n. Now the number minus 1 is not equal to the number less than the distance d from any positive number 8. So if a is greater than 0 as a result no n after which all members of the sequence would be closer to the a than d. Now if a is less than 0 if a is negative then the opposite. We have plus 1 here and here again up to infinity. So no matter what number capital n we choose we will always have number 1 as an element of this particular sequence. And the positive number 1 is on a greater distance from any negative number than 1. Well and finally if a is equal to 0 then obviously it's also not a limit because it's on the same distance from 0 on both sides. Alright so there is no limit that's it. So this is an example of a sequence which does not have a limit. Alright it's just illustrative purposes because you obviously have to understand that there are different sequences. Some have limits, some don't. There might be actually a confusion about certain monotonously increasing or decreasing sequences as the only one which have limit. That's not actually true because I think I mentioned graphically that these are 1, 2, 3, 4, etc. You can always consider these two curves and the numbers being on both sides. So this is a1, this is a2, this is a3, this is a4. They are closer and closer to 0 but they are not monotonously increasing or decreasing. They are jumping around this 0 but still have this particular 0 as a limit. Alright that's again just for illustration. Next, arithmetic progression. Does arithmetic progression have a limit? Well the answer is no and obviously we are adding something to a constant a which means again on this particular graphic if this is a, this is a1, a plus d. Let's say d is this interval. You have a plus d, this is a plus 2g, etc. So the numbers are going to infinity basically. For positive d it goes to a plus infinity. For negative d we are decreasing the game by some number and it goes to negative infinity for negative difference. Well let's just ignore the trivial case of d equals to 0 and then yes, obviously arithmetic progression with difference equal to 0 is basically a constant so it does have a limit. So ignoring that particular case, in all other cases positive or negative difference d. Arithmetic sequence does not have a limit. Now how can I prove it more rigorously than just illustrating this in as a graph and again our purpose is to just teach the technique of how to prove these type of things. Let's go from the definition. Definition is, let's assume that there is some limit. Let's say l is a limit of this sequence. Now it means that l minus a plus n minus 1d, absolute value should be less than, okay there is a small confusion. The difference is d in this case and I used to use d for distance from the limit. So as a distance let's use Greek letter delta. So delta is greater than 0 and this is difference the distance from, this is the distance from the limit. So for any delta greater than 0 I have to find number n such that if n is greater than capital n this is true. Well let's just think about it. For instance for a second that difference is positive for the negative it will be exactly the same thing. So if distance is positive so I really know that the sequence goes to infinity which means if l is its limit then what I'm saying is that l plus delta would be surpassed by adding sufficient number of d. I can prove actually is by finding the number after which it will be surpassed. So it will not be less than delta, it will be greater than delta. How can I find it? Well I will just solve this inequality. This less than a plus n minus 1d. Now d is again positive number. So I can solve this equation how the solution is l plus delta minus a less than n minus 1d. And now since g is positive I can divide both sides and I'm getting n minus 1 greater than l plus delta minus a over g. Where a is the beginning, g is the difference, delta is any distance I choose and l is a proposed limit. So whatever the limit l is proposed if n would be greater than well this fraction plus 1, 1 plus l plus delta minus a over d. So if lowercase n will be greater than this number then I will have this inequality. And this inequality means that the distance between the member of the sequence and the l is greater than delta. It's above l plus delta. So no matter what delta I choose I can find the number after which my sequence would exceed l plus delta. Now if g is negative just a slightly different logic with exactly the same thing. Now if g is negative what I'm saying is I'm going down, I'm making my m members less and less and less. So what I'm saying is eventually my members for negative d would be less than l minus delta. So delta is a positive distance from the l. So if my number is left to the left less than l minus delta it means that the difference between my member of recent progression and the proposed limit l is greater than delta. Now does this inequality has a solution? Well let's try it. m minus 1d less than l minus delta minus a. Now let's remember that d is a negative number which means if I divide both sides by g I should reverse the inequality. So I will get m minus 1 greater than l minus delta minus a over g. And m is again greater than this particular number. So if n is greater than this these are invariant transformations. I will get this inequality which means my recent progression eventually gets less than l minus delta which means it's more than delta distance from the proposed limit l. No matter what l I choose, no matter what delta I choose I will always go beyond that particular limit. So there is no limit. That's my point. Arithmetic progression has no limit if the difference is not zero. How about geometry progression? Well in case of geometric progression you have a, aq, aq square, etc, aq n minus 1, etc. So that's the members. Now let's think about it. No matter what a is if q is greater than 1 you are multiplying some number by a factor which is greater than 1 which means we increase it. So this thing is increasing and we are increasing it more and more and more. So that's why what I can say is that if q is greater than 1 or similarly less than minus 1, absolutely clearly if q is greater than 1 then this geometric progression does not converge. It does not have a limit. If however my q is less than 1 by absolute value let's say it's 1 hat or something like this we are decreasing a every time on multiplying the new member from the previous one by this particular q which is by absolute value less than 1. We are decreasing its absolute value and obviously it should go smaller and smaller down to 0 and 0 is actually a limit. So my proposition here is if absolute value of q is greater than 1 the geometric progression does not converge. If however it is less than 1 it does converge to 0. Now the trivial case when q is equal to 1 is not really interesting because then we will have a, a, a, a, a. It's a constant and limit would be obviously a. Alright so forget about this case. I would also like to forget the negative cases because they are exactly the same as positive so I will do the proof for only positive a and q just to make it easier with all these inequalities and I will prove that if a and q are positive and if q is greater than 1 it diverges. It's not converging but if it's less than 1 it converges to 0. Alright so let's say q is greater than 1. Now what I would like to prove right now that my member of the geometric progression will be greater than any number at certain point whatever I choose as a limit. Let's choose l as a limit and again let's say delta is the distance from the limit and I will prove that it will be greater than this at some point. So I will find the n when this member will be greater than. Now how can I find it? Well very simply. Number 1 I will divide by a and again we have assumed that everything is positive so I don't have to flip the sign. So l plus q over a less than q to the nth degree. Now you should remember if you don't that's okay I'm just telling you that the function called logarithm graphically looks like this log x. Let's say it's like this 10 or like this 2 it doesn't really matter what kind of logarithm we are using. This is x, this is y, this is 0. It looks like this and it's monotonously increasing which means if we will apply logarithm to both sides of inequality we will get similar inequality with logarithm because if monotonous function is applied to inequality stays. So log of l plus delta over a is less than log of q to the nth degree. Now q to the nth degree again that's the property of logarithm. Logarithm of a power is a power times log x. A log q which means I can find n from here by dividing both sides by log q. Now here is very important you see log is positive when my argument is greater than 1 and negative when it's less than 1. Now we can assume that q is greater than 1. So log is positive so I can divide without changing the sign and what I have I have n greater than 1 log over another log. This is a solution. These are invariant transformations which means that as long as my number n is greater than whatever this number is my a q to the nth degree would be greater than l l plus w. No matter what l and w I choose. Which means that my member of the geometric progression is further from any limit than delta which we have chosen. Which means that l cannot be a limit in the story. Alright now what if q is less than 1. Okay if it's less than 1 I claim that 0 is a limit. Alright let's check it out. Let's take any delta greater than 0 and if 0 is a limit I have to find such a number n that if n greater than capital n absolute value of a q to the n should be less than delta. That's what it means because I can put 0 minus delta absolute value but that's also the same thing. And I don't really need absolute value here. No matter how small delta is considering they're all positive we have started from the beginning right a is positive q is positive but less than 1 and delta is positive. So delta is some small number one half one quarter one thousand. It doesn't really matter. But it will be less than this delta after certain n. Now let's find n. Let's just solve this inequality. Again firstly divide by a so q to the n is less than delta over a. Now we will log both of them. Logger is q to the n might make less than logger is delta over a. Now again this is n log q. And I would like to divide by log q right. But log q is negative now because q is in this particular interval and you see if q is in this particular interval it's logarithm is less than 0. It's negative. So dividing by negative I have to reverse the inequality. And what do I have? I have n greater than log delta over a divided by log q. So this is the boundary. So any integer n which is greater than this particular number would fit this particular requirement because if lower case is greater than this then all these transformations are equivalent which means a q would be smaller than delta closer to 0 than any delta. By the way if a is positive and q is from 0 to 1 and delta is very small actually then both logs will be negative. Because this is a very small number delta is supposed to be a small closeness. It's a distance to 0 so we are choosing it as small as possible. And q is negative so if you divide it you will have a positive number. So don't get confused that log q is negative. Log delta over a is also negative. So we have some positive number. And obviously the smaller the delta considering a and q are constant. The smaller the delta the less to the negative side log of this particular number would be. So this is increasing in absolute value while staying negative. And this is negative constant which means that n should actually be greater and greater. The closer the closer we want to get to 0 the greater n we should choose to be that close. Well that's it. I did not really consider negative a's or negative q's. They are exactly the same basically just based all the proofs is based on absolute value and reality. So these are not very difficult but I would say illustrative I think that's the word. Illustrative maybe. Illustrative examples of how to prove that certain number is a limit of certain sequence. Or that certain sequence does not have a limit. It will be a little bit more difficult for the future problems. These are again very simple ones. I do encourage you to do all these problems just by yourself. Go to the website unizord.com and try to solve these problems yourself just to make sure that you understand them fully. Again as always I am encouraging you to register as students to have somebody or maybe yourself register as your students. Again as always I am encouraging you to yourself register as your supervisor or a parent and enroll in certain classes take exams that would quantify your familiarity with all these materials. Thanks very much. That's it for today.