Limits & Continuity Lesson 2

Added: 2026-05-09

[00:00:01]All right. So now let us discuss uh evaluating limits. Let us do the finite limits and set the limits, special limits, everything about limits.

[00:00:14]So remember a limit is a boundary. Yeah.

[00:00:19]So this guy should learn to know his limits or else they'll chop off your head. Yeah. So now um uh this video is a continuation from where the previous one ended from. Yeah.

[00:00:36]So we're now going to look at finite sets at infinity. So this one will just be quite very brief. It is what we just simply look at the behavior of an of an infinity limit. So when x is approaching infinity. Yeah. So uh I I uh I came up with this. So like these we can say they are the rules. Yeah. So these are the basics which you should know about infinity. So like when you divide a number when you divide a number so let's go any number like x when you divide any number by infinity you get a zero. So 1 / infinity is 0. 2 infinity is 0. Just like that. And then for the rest of the operations, if you add a certain number plus infinity, it will continue being infinity. And then if you subtract a certain number by infinity, uh it will continue being infinity. And then if you multiply a number by infinity it will be infinity. Yeah. Just like that. So we're going to evaluate now limits at infinity. Yeah. So taking into consideration of what we've just listed out. So uh this uh this this is an example I'm going to have dealing with uh the infinity. So let's compute the limit as x approaches infinity 5 + 1 /x. So remember we just substitute. So when we substitute we're not going to bring down the limit as x approaches infinity. This is 1 / uh infinity. So 5 + what is a number divided by infinity? Uh like we have said it's zero and then 5 + 0 that's five. So you can see this as an MCQ. Yeah, you can see it as a multiplechoice question. Yeah. And then the other one the limit as x approaches infinity 12 pi <unk>3 and then infinity squared. Oh and I also forgot. So if you add infinity and infinity it will be infinity. If you multiply infinity by infinity it is infinity. Yeah. It is just the same like that. And then if you divide infinity by infinity it is an intermediate form.

[00:03:38]This is not equal to one. It will just be like that. We can even say is infinity.

[00:03:46]Yeah. So this is when we square. So have three uh 3 pi roo<unk>3. So infinity by infinity it is just infinity and then anything divided by infinity it is zero. Yeah. So this is how we go about it. So just take note of these basics I've taught you. Uh they'll be able to help you to answer almost uh 80% yeah of questions that will deal with this. So uh just a quick recap uh infinity plus any number it is infinity and then infinity minus any number then the number it will just go on decreasing and decreasing negative infinity and then infinity times any number this will be So infinity * any number it is infinity. And then if you divide any number by infinity it is zero. Yeah.

[00:05:01]Then your question would be what about if we have infinity over any number this will be zero. Yeah. And then also take note of these then we are good to go. Yeah. So now uh let's go to to discussing the other part. Yeah. So uh let us talk about sided limits.

[00:05:34]Uh uh one-sided limits. What should we know about them? Uh oh. But before we can uh go to to sided limits uh it would be it would be better if I reach the special limits. Yeah. So let us talk about the special limits and then we'll go to the other one. Yeah. So almost done. Special limits. uh what what's special about certain limits? So I have a lot of special limits but some of them are not really special. Yeah. Uh so these are the special limits that exist. They are even more than this. So uh what what is there is that in the when you open a textbook you find a lot more. There'll be number of other limits and then in the lecture slide they only list this and then here we only list these two and then in the exam or in the test you only see this in the exam they'll bring that yeah and then for these uh I do not pay much attention to them but you can just study them for safety in case they it comes as multiple choice. When this happens, it will be equal to what? You have to write = 1.

[00:07:08]The limit as x approaches zero. Uh the exponent x - 1 /x. Yeah. But for this we'll come to it when we go further in calculus. Yeah. We we apply this somewhere and a little bit of this.

[00:07:31]So let us now talk about the special limits. So the the limits which are really special are these ones this and that. So the limit as x approaches 0 of sin x /x = 1. Uh what does this mean?

[00:07:50]So uh what we're saying the first one is that the limit as x approaches is it zero sin x / x = 1. Uh and for this one you might argue like okay but when we substitute x here it will be zero. This will be infinite.

[00:08:20]This will be undefined. So I think this is the part where now this guy oversteps the boundary. He oversteps the boundary and faces his fears. He will just be like ah no today I'll try my luck and then he proposes they reject him. He goes away.

[00:08:42]Yeah. So now uh talking about limits. So this one uh we don't really substitute. There is a way of proving but I don't want to prove that we just waste so much time.

[00:08:58]Yeah. So we just know that the limit when x is approaching 0 sin x /x = 1. So this is true even when you have something like sin uh even when you have x x / sin x the limit as x approaches zero x / sin x yeah this you can just conclude that it's equal to one as well yeah so this is true even when you have like uh here you have ax x uh here you have a x over a x. So as long as uh what is here is exactly the same as what is there then the limit is equal to one. Yeah. So like here you can see we have ax even here we have ax then uh this will also equal one. Yeah.

[00:09:58]So we're going to look at a few questions. The aim just to make sure that here and there it's the same. As soon as it is same then uh we're going to to finish it up.

[00:10:14]Yeah. So uh that is the first part and then uh what is the other part which we should know?

[00:10:25]So many questions will come from this first one and then uh a number of other questions as well. They'll come from the other uh from this other limit x approaches uh is it zero. So I have 1 - cos x over uh / x 1 - cos x /x. So this will be zero. So where wherever you see the whole lot of this just write it that it's equal to zero. It's a nobody. Yeah. So now let's get evaluating. Uh how will questions come? Uh so we can have an example. What what example can I give?

[00:11:26]So let's say you are told to find the limit as x approaches zero. And then we have sin uh sin 3x / x. Yeah. So what can you do when you have such? So here you have to remember that okay when I have the limit as x approaches zero let me write that better. So when I have the limit as x approaches zero uh sin x over x the whole load of this is one. Yeah. So this is our key. So remember I said as long as the thing here is the same as that then everything equals one. So like here we have 3x but here we have just x. So in such a case this is not yet equal to one. We have to make sure that even here three that's when it will be equal to one. So I've even given you a hint. Uh so what we need to do is we multiply by three but we can't just multiply down by three.

[00:12:48]Even on top we should multiply by three.

[00:12:51]Yeah. So what is 3 * this? So we already have a three here. We do not want to have a nine then the things will be different. So three times the the things the whole lot of this we can take it to be in front. So on the rules of limits we have the uh the constant row. Yeah. So when you're multiplying a limit it simply means the number is like outside t.

[00:13:21]Yeah. So the three goes out there and then this three we can also write it here as well but we do not write we do not want to write it there. So sin 3x over x. So the three we can it can come here so that we have 3 over 3. So we haven't changed anything because 3 over 3 is one and yeah. So but since we're not going to take it that side, we're just going to bring it here. Yeah. And then in this case we have a 3x. Yeah. So since we have 3x and 3x it means that we have 3 * the whole lot of this. Since we have the same thing on top and below, then the whole lot of this is 1. So 3 * 1 is 3. That's how you evaluate uh such limits. Yeah. So can we do another one?

[00:14:32]Um what example can we think of? Yeah. So now let's think of a case where this is a lecturer's favorite where we're going to combine. So like x is approaching zero and then we have sin uh sin 5x over uh what can I put over cos cos anything cos 2x like that. So uh in in this case then we would first have to to separate uh yeah so we'll separate meaning we're going to have the limit as x approaches zero. So we have sin 5x * 1 / cos 2x. So usually for co we do not have to worry much because uh even if x is approaching zero uh cos zero it is just one. Yeah.

[00:15:58]So here I've separated the two. Yeah. So like we can multiply back one * this. It is that same thing and then cos 2x * here it will be cos 2x. So this simply means that here we have an over one. Yeah. So I've said for coh there are no problems. We're just going to substitute where there is x we put a zero. Yeah. But for sign if we put a zero then sin zero uh we're going to to get something we do not want. So what we're going to do is since we have a 5x here what do you think we should have here? Think think think. Yeah. So what should we have since we have a 5x? So obviously we should have a 5x as well. So multiply throughout by 5x. Yeah. So I'm multiplying by 5x because we can cancel it out. So we haven't changed the meaning. Yeah. So 5x * this. We already have a 5x here meaning it will go in front. So we're going to have 5x there. It's in front. And then we have limit as x approaches zero.

[00:17:20]So the limit as x approaches zero sin 5x over one. So this 5x which is down if we want we can bring it here. Yeah we can bring it here but we'll just put it here so that they can match times. So this limit it is for everything here. So we get * 1 / cos cos 2x. Yeah. So at this point we're going to say oh okay so we'll simplify first will distribute this limit. Remember it is for both of these. Yeah. So we're going to write the limit twice so that when we simplify we do not have problems. So we have a 5x outside and then the limit as x approaches zero sin 5x over 5x times again the limit as x approaches 0 1 / cos 2x X. Yeah. So the whole note of this the limit as X approaches zero 5 X and 5X meaning so this is the five is outside times the whole lot of this is one. Yeah. Times so the limit as X approaches zero. So remember when we substitute what is X when we put zero then we stop writing down the limit. So let us substitute this will be 1 / cos 2 0. Yeah. So I hope you haven't gotten so confused. So this is 5x * 1 * uh 1 / 1. Yeah. Cos so this is 2 * 0 0 cos cos 0 it is 1. Yeah. And then the final answer will just remain 5x. Yeah. So this is 1 * 1 * 5. Yeah, that's how you you work it out.

[00:19:54]Yeah. So let's do just Oh, I haven't yet done anything relating to this. The limit as x approaches 0 1 - cos x over x. Yeah. So when you see that you're having something that is mainly relating with co then if we are going to have the same denominator. Yeah. So usually here we just have cos x we say that that part is zero. Yeah.

[00:20:30]So let's have an example. Okay. So this is what we have.

[00:20:38]Yeah. So I just want to do something relating to this one here. Uh okay. So here we're going to do uh this question. Yeah. So so this uh this special limit is not very commonly used. Yeah. But I'll look for more questions on this and then we'll handle them. So how can you work out this one? We have been given that limit. So we can rewrite this. We can say we have the limit as x approaches 0 -1 + uh + cos x over x 2. Yeah. So that at least it now looks like that one. So what we're going to do now is we factoriize. So uh let's factoriize we say the limit as x approaches zero. So here when we factor out a1 we're going to remain with so fac with one. Yeah because 1 -1 * this it will take us back to the original one. So we factored out -1 and then this positive we factored out1 it's it remains negative and then we have cos cos x. So this is over x². Yeah. So here we've just brought out the negative outside. Yeah. In simple terms we've brought out the negative outside. we've written -1. So meaning here the negative has gone away and this one has become uh the positive has become negative. Yeah. So that even if we multiply them back negative and negative it will be positive. Yeah. So we factored out the negative and then we can now factoriize further. So this1 if if we have a constant in the function we can take it outside. Yeah.

[00:23:01]So this will be1 we've taken it out over. So we have two x's here we're going to take out one as well. Yeah. So, and then we remain with the limit as x approaches zero 1 cos x over uh x here just remain with one. We factored out the other one. So at this point it looks like uh the the general one which is right there which is here 1 - cos x over cos x. So the whole lot of this according to what we know the whole lot of this is zero. So time 0 and anything time z is zero. Yeah. So this is how you can work it out. So usually when you have something dealing with cos x - one, you're just going to end up with a zero somewhere. Yeah. So we'll work out on more examples. Yeah. But not in this video.

[00:24:16]So the these others you can multiply by the conjugate or the numerator. Here you just put a plus. You multiply throughout. Here you you simplify the numerator. The numerator is a fraction.

[00:24:29]It has 1 / 2x or 1 / 2 + x. Uh it does 1 / 2 + x - 1 / 2 and then it's over x. So you simplify this, you make it as a single fraction and then you divide by x. So you can try it out. And uh whenever you get stuck uh you are always free to to inbox me. Yeah. And then this one uh it simply means that down here we should also have a three. So meaning we multiply throughout by three. So meaning this three will go outside. Yeah. And then we have limit x approaches zero. And then we have sin 3t over 3t. This one has come here. And then this one I've taken it outside. So the whole lot of this is 1. So 3 * 1 3.

[00:25:33]Yeah. And then uh and all all these. So you just have to have a starting point here. You just factor out one x. 1 x will go outside.

[00:25:46]And then you're going to have this sin x / x. Yeah. So the whole lot of this is one. So the final answer here is one. Yeah. And then here you try to rationalize the numerator. You multiply by the conjugate. You see why you can cancel if possible. Yeah. You're just moving like that. And then this satanic one.

[00:26:12]Uh try simplifying it. Yeah. So when you have a cube root, it simply means to the power 3. Yeah. So you have to the power 3. You can multiply both sides by x^ 3. Even there x^ 3. So that this and that will cancel. Uh yeah. So first you try and then you come like okay here things are not so well for this one. Quite funny.

[00:26:43]uh 32 can be raised to the^ 5. So minus 2^ 5. Yeah. So we can open x - 2 x - 2 x + 2 5 times. Yeah. And then you cancel out with this. Yeah. So the key thing is just to remember that 32^ 5. Yeah. So uh I don't want to go into too much detail.

[00:27:14]Yeah, you you've learned the basics. So you should try practicing on as much as other things as you can. So now let's make progress uh uh let's do uh one-sided limits now. So uh what do we have to know about one-sided limits?

[00:27:37]Uh so one-sided limits uh so I'm sure some people feel attacked like uh uh when people attack us in one-sided relationships uh so the onesided limits they are [Music] simply so we have a function so you know functions like a parabola We have a root of a function. It looks like that. We have uh yeah so you know how to plot different kinds of functions. So one-sided limits we normally we analyze them using the cartian plane. Yeah. So here where do I start from the one-sided limits? Okay.

[00:28:34]So let's say want to plot the graph of y is = y = what 1 / x for example. Now this one it won't make sense for a start. Uh what can I think of? Uh my my my idea is gone. Yeah.

[00:29:04]So uh the one-sided limit it is by take any form. So like this is the graph of y is = root x. Yeah. So now when we're talking about one-sided limits, it's simply a part where we look at a limit uh from a specific point of view. Yeah.

[00:29:29]So this uh popular social media abbreviation.

[00:29:34]Yeah. So we look at a limit from a particular point of view. So like when x is approaching zero, where is it coming from? Are we looking at it from the negative side or from the positive side?

[00:29:49]Yeah, that's one-sided limits. Ah so now like in this case we can have a limit where by uh I just couldn't think too much I don't want to stress myself so like let's say you have a function uh we're going to say so this is the graph of root x so now what if we have the graph of root x uh + 2 yeah so the graph of root x + 2 or we can write minus2. The graph of x - 2 from functions part one what we looked at in semester one. Uh oh that was graphs of functions. So it would be like somewhere here. Yeah. So I just doing some rough work. Yeah. It will so this is 0 1 2. It starts from here. it it goes there.

[00:30:51]Yeah. So there are values whereby this function it can be complex like for example values which will make it to be negative. Let's say if x is 1 if x is one or I've written negative if x is 1 then it means we're going to have 1 - 2 which would be equal to -1. So this is an imaginary number. So like if I to plot we would say that uh this is the graph of uh <unk>x - 2 but it's going to have uh so like this is the point where x is one uh in the y- axis. So like would draw a dot. Yeah. So a dot simply means not part of this. solution. Yeah. So this is not my best example. Yeah. So that is just uh a mere introduction. Yeah. So inside one-sided limits it's whereby like we can have a function it is continuous but there's a removable discontinuity. Yeah. Like if there's a dot then that's not part of the solution. Yeah. So but if you have a function it is just continuous it means this is all all the values are in the domain.

[00:32:22]Yeah but when there's a dot it means the values are not part of the domain. Yeah. So now how do you evaluate one-sided limits? So for the one-sided limits like I said we normally just look at we'll be dealing with the x or y plane. Yeah. So let's go straight to an example. Uh okay. So before we do an example uh firstly we have to understand uh what qualifies a limit to be a limit.

[00:32:59]So there are three conditions that must be satisfied. So if we have a function like f ofx and then want to determine we're looking at the limit when x approaches a certain number like maybe a. So uh the the conditions are that when we approach this limit from uh from the negative side of a when we approach it from the negative side of a it should give us a certain number like it should give us b and then when we approach it uh when we approach the same limit from uh the positive side when we approach the same function from the positive side it should give us uh B again. Yeah. And then also finally when we just substitute so like when we just substitute where we say f of a so meaning in the function where there's a when we substitute again we should still get b. So the limit has uh too much restrictions. Yeah. So let let me put you into perspective with this example. Like for for example if we have f ofx is equal to 1 / 1 / 2x. So if want to determine whether the limit when the limit as x approaches uh one if if want to determine whether this limit exist it it has to satisfy this condition the first one is that when you approach one from the negative side it should give us a particular number. Yeah. So like let's just quickly do it. So uh one you know how Yeah. Okay. So this is 0 1 2 the numbers are just going like that. So when we approach one from the negative side meaning like we can get 0.9. So here is x what happens when we put 0.9. So it means we're going to say uh 1 / 2 * 0.9 and the answer is [Music] uh is 0.5 0.55 so or 0.56 we've approached the limit from the left but what if we approach it from the right like so from the right it means we're getting a number greater than and we can get 1.5 1.1. So what happens when we get 1.1 then we're going to have 2 * 1.1 and then you divide it you divide one into it. So here when you put 1.1 the answer you are getting is 0.45. So you can see that when you approach from the left and when you approach from the right you're not getting the same thing. The other one is 0.45. The other one is 0.55. So the limit will not exist for such. Yeah. And then even for the third condition, it simply means we directly substitute. Yeah. So like here if we put one it means 1 / 2 * 1 is 2. So 1 / 2.

[00:36:45]So when we put one here, the final answer is 1 / two. So the the answers are different when you get from the negative side when you get from the right hand side the positive side you're not getting the same things. Yeah. So we do not usually prove like this because it would require you to use a calculator and mathematics at a level doesn't require calculators. Yeah. So this is just for explanation. Uh yeah. So but um in case the lecturer just wants to punish you, you did something bad, they'll ask you to prove without giving you a graph.

[00:37:23]Yeah. So but when we're given a graph, that's when it becomes way too easy. Yeah. And the graph is confusing only when when you haven't learned it. Yeah. So but uh let's make it all simple.

[00:37:44]So here the the the question is find the limits of the function graphed. So the whole lot of this is one function. Now it has uh it has se it has segments.

[00:37:58]Yeah it has uh discontinuities like it has started from here instead of just going straight it goes down and then it continues. It goes up it goes down. So this is just one function. Yeah. So we're going to find the limits at at each point. So the first point zero, the the second point 1 2 3 4.

[00:38:25]We're going to find the limits four limits.

[00:38:30]Yeah. So this is a typical exam question. Yeah. And then we're also going to look at an exam question shortly. Yeah. So uh where do I start from? So let us look at the point where x is zero. Does the limit here exist? So for the limit to exist it means that if we follow the limit as x approaches zero from the left hand side it should be the same when we approach the limit as x approaches zero from the right hand side. So here we just follow the graph like okay zero is here from the left we do not have any graph. So on the left h limit does not exist. The limit does not exist here on the left we don't have any graph. Yeah. And then what about on the right? So on the right we now want to come to zero meaning we're coming in this direction. Yeah. So we're going to start from the uh like anywhere like we're starting from here. We're going we're going until we reach the point where the W is zero which is there. So when we're approaching zero from the right the value which we have it is one. Yeah. So this is the value which we have. Yeah. So I think uh for now it doesn't made so much sense but just hold on. Yeah. So from the right it's very easy to follow through. We're follow the graph is we're following it like this but it just fell down and then it it is taking us to one.

[00:40:26]So from the right it is one from the left it does not exist. meaning um the the limit in itself just does not exist in this case. Yeah. So the limit when x approaches zero does not exist because from the right and from the left are not the same unless if they are the same. If it's one even there it's one then it does not exist. Yeah.

[00:40:53]So uh let's go to the next point which which will clarify things better. Uh so at the point one what happens? So let us approach the limit when x goes to 1 from the negative and when and the limit when x goes to 1 from the positive. So here uh from the negative you know the negative we go like this. Yeah. So we're coming down to one. So when we reach uh the point one is the value which we have we have a zero. The y value we have a zero. So when we're approaching one when approaching one from the negative we have a zero.

[00:41:49]And then now what if we're approaching one from the positive uh from the positive? Yeah. So we're going to come from here up to we find one one it is uh right there. So this is the uh the value of one. Yeah the x value of one. So we've reached this point. So what y value is it going with? It is going with positive one there.

[00:42:28]Yeah. Okay. Yeah. So uh uh we found from the left when we came from the left it is zero and then when we're coming from the right now it is at one.

[00:42:49]Yeah. So since from the left it's zero from the right it's one then the final limit it does not exist. Yeah. So we conclude therefore the limit when x is approaching one does not exist. Yeah. So uh let's uh practice on another one. So we've seen so both of these points there on one it depends on where we are coming from.

[00:43:22]When you're coming from the left the graph was coming down and then when you're coming from the right it was going like this.

[00:43:29]Yeah. So let's go to two. Uh this point we're going to approach it from the left and from the right. Uh so what are we going to to have? So when we approach the limit when uh when x approaches two from the left and when approaching two from the right uh what should we expect? So here what we're simply going to have is so this is two when approaching it from the left the graph is coming it's coming it's coming we have reached two. So what is the y value of two at this point? When you look at it, it is one.

[00:44:19]It is one. Yeah. And then let us approach two from the right hand side. Uh from the right hand side, we're coming like this. We're coming to two. So it was coming down. So from the right hand side when you look at it is also going hand in hand with one. Yeah. Oh yeah. Okay. This is so nice. The left and the right they are both one. When you approach it from the left it is at one. When you approach it from the right it is also at one. Yeah.

[00:44:58]So but we're not yet done. So what about um so remember there are three functions left right and also f of a so meaning here we simply mean f of two um f of two so what is the value of the function when f is two so remember I said when you have a dot a dot means it's not part of the solution yeah the point is just there but it's not part to the solution. Yeah. So like in this case uh you can see that on two we have two points. We have two dots have this one and that one. So the one which is for f of two it is the one which is shaded. So f of two it is going with two. So we have three different things or have different things meaning the limit does not exist again. Yeah. So at least this one it tried it was almost there. So the limit you conclude. So you have to conclude the limit when x is approaching 2 does not exist because uh it is dotted here unless uh it is not shaded here. Yeah. So unless if it was shaded then if here it was also one then it could have been true. Yeah. So let us uh do another one. Yeah. So by the end of the last example you you will have understood everything. So uh what about the point three? Uh what happens here at this point? Let us So we're now going to three. The limit as x approaches 3 three. Yeah. So now here we're going to approach it from the left. We approach it from the right. And then we check f of three. What is the dotted point?

[00:47:06]Yeah. So uh we can uh we can go through it. So from the left three is here or and you can even draw a line to know where you are coming like you want to approach the line for three. Yeah. So we'll start from the left. Uh the graph is going you you can start from any as long as you are going to as long as you coming from the left. So we've reached the point it is two when we're coming from the left. What if we're coming from the right it will go? Yeah it has reached. So when it's coming from the right the point is uh also two. And then what is the dotted line at three? This is what it means f of three. What is the dotted line at three? The dotted line at three. Oh, not dotted the shaded. So if another point like this, we cannot consider this one because it is it is not shaded. So the shaded point at three has a y value of two. So you can see have 2 2. So it means that the limit as x approaches 3 exists and it is equal to two because from the left it is two from the right it is two and also the f of three it is two again. Yeah. So this one it satisfies all the condition. Yeah. And then number four uh oh battery low. Okay. So number four, we're going to say uh let's quickly do it. So number four from from the left, this is the left hand side. The left always go like that. So from the left it is at one. So from the left hand side it is at one. What about from the right? There's no graph on the right of four. So it does not exist on the right. And then f of4 where is it dotted? It is dotted at uh 0.5. So you can see these values they are not uniform. They have to be the same. So therefore this limit does not exist. Yeah. So now we're going to uh go to an exam question. Yeah.

[00:49:51]So like I said even the one was just from analyzing it's actually exam material. Yeah. Although maybe they can make some few changes. So this is how the exam will look and then if you didn't understand well you'll be like ah that was so bad an exam. Uh yeah. So uh unfortunately uh this lesson will end here.

[00:50:21]uh due to power issues. So in the next video the video you'll find it on the link or on the comments. So just check on the comment of this video you're going to find the continuation. So it will be very short.