WEBVTT 00:00:01.001 --> 00:00:08.200 If you haven't already done so you should review topic 1, limits from a graph, before starting this topic. 00:00:08.202 --> 00:00:13.100 Let's take a look now at limits from an equation. 00:00:13.102 --> 00:00:17.000 We should review the working definition that we've been using for limits. 00:00:17.002 --> 00:00:24.501 The limit as x approaches c of f of x equals the single real number L if and only if 00:00:24.503 --> 00:00:28.900 the limit as x approaches c from the left of the function is L 00:00:28.902 --> 00:00:34.800 and the limit as x approaches c from the right of the function is L. 00:00:34.802 --> 00:00:40.090 So these limits must both exist, and they must equal the single real number. 00:00:40.092 --> 00:00:48.200 Now interpreting this is a little different when all you have is the equation. Let's take a look at several examples. 00:00:48.202 --> 00:01:06.000 Example 1. Let's find the limit as x approaches 1 of the function 2x plus 3. 00:01:06.002 --> 00:01:12.701 So without the sketch to read from, we need to analyze the function. Let's look at what we have. 00:01:12.703 --> 00:01:21.300 We have 2x plus 3, so this is the f of x. Now if you recall, this is a linear function. The graph is a line. 00:01:21.302 --> 00:01:27.700 It has a slope of 2. It has a y-intercept of (0, 3). Now if you think about what you know about lines, 00:01:27.702 --> 00:01:32.400 unless there's anything peculiar to affect the domain, the domain is going to be all real numbers. 00:01:32.402 --> 00:01:38.300 The function is going to have no holes, no asymptotes. And so if you picture this, 00:01:38.302 --> 00:01:42.000 you aren't actually sketching the graph, but you can picture what it looks like. 00:01:42.002 --> 00:01:48.200 You know that on a line the limit from the left and the limit from the right are going to be the same, and 00:01:48.202 --> 00:01:55.001 they're going to be the value of the function at the point. This is a polynomial function, and you will 00:01:55.003 --> 00:01:59.300 realize over time as you practice this that when this function is polynomial, 00:01:59.302 --> 00:02:06.401 all you actually have to do is find f of 1. That will be the limit from the left, the limit from the right, 00:02:06.403 --> 00:02:11.801 the limit, and the value of the function at the point. So we call this direct substitution. 00:02:11.803 --> 00:02:18.801 We take the 1, we substitute it into here, we get 2 times 1 plus 3, and so this limit is 5. 00:02:18.803 --> 00:02:27.101 Again, we can do this direct substitution because we analyzed the function. Let's take another one. 00:02:27.103 --> 00:02:39.101 Example 2. Let's look at the limit as x approaches negative 3, and let's use a quadratic function, 00:02:39.103 --> 00:02:49.700 say x squared plus x minus 1. Again, no graph to read from the left and the right, so we analyze the function. 00:02:49.702 --> 00:02:54.200 This is a quadratic function. The sketch is a parabola with no restriction on the domain. 00:02:54.202 --> 00:02:59.800 It's a smooth curve. The limit from the left and the limit from the right are going to be the limit 00:02:59.802 --> 00:03:06.401 and the value of the function at any point. So again, because of that we can use direct substitution. 00:03:06.403 --> 00:03:15.600 We substitute negative 3 into here, so we get negative 3, squared, plus negative 3, minus 1. 00:03:15.600 --> 00:03:24.600 So that would be 9 minus 4 which is 5. Alright, let's take a look at another one. 00:03:24.602 --> 00:03:40.101 This one will be a constant function. The limit as x approaches 2 of 3. Now this looks a little different 00:03:40.103 --> 00:03:47.601 because there is no x. So let's think about the function, analyze it. This would be f of x equals 3, or 00:03:47.603 --> 00:03:54.300 y equals 3. The graph is a horizontal line. It has no holes, no asymptotes unless there's something 00:03:54.302 --> 00:04:00.201 peculiar about it, which there's not in this case. So the limit from the left and the limit from the right 00:04:00.203 --> 00:04:06.401 and the limit and the value of the function will always be 3. Now it looks like here 00:04:06.403 --> 00:04:13.002 that you're just sort of copying this number over from here to here, but what you're actually doing is 00:04:13.004 --> 00:04:19.701 you're analyzing the situation, and if you want to you could always just get a rough sketch, 00:04:19.703 --> 00:04:30.300 just this horizontal line. You can see that at 2, right here at (2, 3), the limit from the left and the limit 00:04:30.302 --> 00:04:37.500 from the right and the limit are all going to be 3. Now let's take a look at some piecewise graphs. 00:04:37.502 --> 00:04:43.801 Example 4. Let's take a look at a piecewise function this time. We want to find the limit 00:04:43.803 --> 00:04:51.900 as x approaches negative 1 of the function f of x, when... and then here's the definition of f of x. 00:04:51.902 --> 00:04:59.000 This function is defined in three pieces. We have one piece where the domain is values of x less than 00:04:59.002 --> 00:05:07.000 negative 1, another piece where the domain is just at the point where x equals negative 1, 00:05:07.002 --> 00:05:13.001 and then the third piece where x is greater than negative 1. Now even though we don't have a sketch 00:05:13.003 --> 00:05:19.700 here, we can visualize what's happening. We can find the limit as x approaches negative 1 from the left 00:05:19.702 --> 00:05:24.700 by using this piece of the function because these are the values less than negative 1. 00:05:24.702 --> 00:05:30.398 We are unconcerned with this piece because this tells us what happens at negative 1, 00:05:30.400 --> 00:05:32.300 and that has nothing to do with the limit. 00:05:32.302 --> 00:05:39.700 Then we look at the limit from the right using this piece because here x is greater than negative 1. 00:05:39.702 --> 00:05:51.300 So let's write this down. We would find the limit as x approaches negative 1 from the left. 00:05:51.302 --> 00:05:55.801 Now I'm going to stop here just a minute and point out something. This looks a little strange 00:05:55.803 --> 00:06:01.701 the first time you see it. This means negative 1 from the left. 00:06:01.703 --> 00:06:07.501 Alright, and of the function f of x, but in this case the piece we're looking at is x, 00:06:07.503 --> 00:06:13.101 so I'm going to just use that piece because that's where the domain is less than negative 1. 00:06:13.103 --> 00:06:20.100 So I'm looking for the limit of this function as x approaches negative 1 from the left, and because 00:06:20.102 --> 00:06:29.600 this is a polynomial function, I can use direct substitution. I sub in negative 1, and there's the limit. 00:06:29.602 --> 00:06:41.200 Ok, that's the a part. Let's look at the b part. What is the limit as x approaches negative 1 from the right? 00:06:41.202 --> 00:06:48.098 Well, that's this part of the domain, so I'm going to use x plus 2, again a polynomial function. 00:06:48.100 --> 00:06:57.101 This is a line, just as this is. I can use direct substitution. I sub in negative 1. Negative 1 plus 2 is 1. 00:06:57.103 --> 00:06:58.600 So what do I have? 00:06:58.602 --> 00:07:04.199 Well, I have a situation where the limit from the left does not equal the limit from the right, 00:07:04.201 --> 00:07:10.700 and so you know what that means. That means the limit as x approaches negative 1, and here I go 00:07:10.702 --> 00:07:19.898 back to writing f of x because I'm using the combined function, and I put does not exist. 00:07:19.900 --> 00:07:27.500 Ok, example 5, another piecewise function, but this time we want the limit as x approaches 2. 00:07:27.502 --> 00:07:32.501 Now let's analyze the function and the situation we have here with the domain. Notice that 00:07:32.503 --> 00:07:40.299 this piece is defined where x is less than 2, and this piece is defined where x is greater than 2. 00:07:40.301 --> 00:07:45.404 That means the function is not defined at x equals 2 at all, but that's ok because that 00:07:45.406 --> 00:07:50.310 does not affect the limit. We're going to read the limit from the left with this first piece 00:07:50.312 --> 00:07:56.201 and the limit from the right with this second piece. So let's see what we have. 00:07:56.203 --> 00:08:08.403 We have the limit as x approaches 2 from the left, so I'm going to use this piece. 00:08:08.405 --> 00:08:14.705 This is a polynomial function. The graph is actually a parabola, just the basic x-squared shifted up 1, 00:08:14.707 --> 00:08:21.904 which means the limit as x approaches 2 from the left will be the actual value of this function at 2. 00:08:21.906 --> 00:08:30.500 I sub 2 into here, and I get 5, and again I could do that because this is a polynomial function. 00:08:30.502 --> 00:08:39.203 b. The limit as x approaches 2 from the right, and this time I use the piece where we define 00:08:39.205 --> 00:08:54.101 the domain to be x greater than 2. So I use this piece, and I substitute in here 2, and I get 5. Good. 00:08:54.103 --> 00:08:59.000 So what we discover is the limit as x approaches 2 from the left and the limit as x approaches 2 00:08:59.002 --> 00:09:12.505 from the right are both 5. So, that means the limit as x approaches 2 of the entire function, f of x, is 5. 00:09:12.507 --> 00:09:23.502 And again, let's point out that f of 2 actually does not exist, but this has no affect on the [limit]. 00:09:23.504 --> 00:09:30.101 Alright, let's look at example 6. We have here the limit as x approaches negative 2 of 00:09:30.103 --> 00:09:35.604 x-squared minus 4 over x plus 2. Now this is not a polynomial function. This is a 00:09:35.606 --> 00:09:42.002 rational function, and so we notice right away that if we tried to sub negative 2 into 00:09:42.004 --> 00:09:48.103 the denominator we get a zero denominator, and of course division by zero is undefined. 00:09:48.105 --> 00:09:54.501 So we have to think of some other way to do this [limit]. Now there are a couple of possibilities. 00:09:54.503 --> 00:10:00.901 The one I would suggest first is to try algebraic simplification. This is a tool 00:10:00.903 --> 00:10:04.804 that we're going to use later in another topic, so it's a good one to practice here. 00:10:04.806 --> 00:10:12.505 Let's take a look at this function. We see that if we take the x-squared minus 4, we could 00:10:12.507 --> 00:10:25.501 factor that as x minus 2 times x plus 2, all over x plus 2. As long as we realize that 00:10:25.503 --> 00:10:34.700 x cannot equal negative 2, which is fine, that does not affect our limit, we can then 00:10:34.702 --> 00:10:43.502 cancel this factor. We get 1 here, 1 here, multiply across, and we get x minus 2. 00:10:43.504 --> 00:10:48.900 What we're going to do now is we're going to do basically a new limit. We're going to 00:10:48.902 --> 00:10:58.905 find the limit as x approaches negative 2 of x minus 2, keeping in mind this restriction. 00:10:58.907 --> 00:11:05.605 We now have a polynomial function. I can directly substitute, and I get negative 4. 00:11:05.607 --> 00:11:10.000 This is a technique that you always try to use. It doesn't work for every function, but 00:11:10.002 --> 00:11:16.201 when it does you use it. Ok, let's take a look at example 7. We have another situation 00:11:16.203 --> 00:11:20.903 here where we have a function that's not a polynomial function, and if we did try 00:11:20.905 --> 00:11:25.303 direct substitution, we'd get 9 minus 9 which is 0. We'd have division by zero again, 00:11:25.305 --> 00:11:30.403 and so that's not going to help. So what we do again is we try algebraic simplification. 00:11:30.405 --> 00:11:36.600 Well you notice there are some similarities here because x is the square root of x squared 00:11:36.602 --> 00:11:42.604 and 9 is 3 squared. So maybe there's something we can do to rearrange this function and 00:11:42.606 --> 00:11:48.805 to eliminate the x minus 9 in the denominator. So let's give it a try. Ok, we pull the 00:11:48.807 --> 00:11:59.800 function out, so we have the square root of x minus 3 over x minus 9. Well, what can we do? 00:11:59.802 --> 00:12:05.607 We want to get x minus 9 in the numerator so it will cancel. So we do what's called 00:12:05.609 --> 00:12:15.001 rationalizing the numerator, and we multiply by the conjugate. So I'm going to multiply 00:12:15.003 --> 00:12:23.003 this term, numerator and denominator, by the square root of x plus 3 [over] 00:12:23.005 --> 00:12:27.002 the square root of x plus 3 which of course is 1. It doesn't change the value 00:12:27.004 --> 00:12:32.212 of the function. It just changes the look of it. So what do we get? 00:12:32.214 --> 00:12:37.701 Well we get the square root of x times the square root of x is x. 00:12:37.703 --> 00:12:42.903 We get middle terms which cancel out, which is exactly why we did this. 00:12:42.905 --> 00:12:47.005 We get negative 3 square root x [plus] 3 square root x, which is 0, and then 00:12:47.007 --> 00:12:54.005 we get negative 3 times 3, which is negative 9. That achieves our goal. 00:12:54.007 --> 00:12:58.304 Now the problem we have here is we still have these two terms to deal with. 00:12:58.306 --> 00:13:03.903 Don't multiply this out. Just copy this over right here. Now what we get, 00:13:03.905 --> 00:13:14.103 I'll rewrite it down here, is we get x minus 9 over x minus 9, times the square root of x 00:13:14.105 --> 00:13:21.004 plus 3. Let's see, did we make any progress? Yes we did. 00:13:21.006 --> 00:13:25.605 This factor can be cancelled. We get 1 over 1. Now we need to be careful. That gives us 00:13:25.607 --> 00:13:35.000 a restriction, x cannot equal 9, but that's ok. That's not going to affect our limit anyway. 00:13:35.002 --> 00:13:37.203 We're interested in the limit from the left and the limit from the right, 00:13:37.205 --> 00:13:43.800 not what happens at 9. So what do we actually have left? Well we actually have left 00:13:43.802 --> 00:13:54.000 1 over the square root of x plus 3. So what we're going to do now is we're going to take 00:13:54.002 --> 00:13:59.901 the limit as x approaches 9 of this function. Now even though again it's not polynomial, 00:13:59.903 --> 00:14:08.702 when we substitute in the 9 here, we're going to take the limit as x approaches 9. When we sub 00:14:08.704 --> 00:14:20.800 in the 9, we get the square root of 9 is 3. 3 plus 3 is 6, and so that is going to be our limit. 00:14:20.802 --> 00:14:25.003 Alright, let's take a look at something that doesn't simplify algebraically. 00:14:25.005 --> 00:14:31.603 In this example, we're going to take the limit as x approaches 2 from the left. 00:14:31.605 --> 00:14:38.104 The function we're dealing with is 1 over x minus 2. Now this function is not polynomial, and 00:14:38.106 --> 00:14:43.502 there is going to be no way to simplify this algebraically, so direct substitution is not 00:14:43.504 --> 00:14:48.003 going to be a possibility here. There are some other possibilities though. We could sketch 00:14:48.005 --> 00:14:54.001 the graph and read from our sketch, or we could make a table of values and look at it that way. 00:14:54.003 --> 00:15:05.102 First, let's see if we can sketch this graph. Now you may remember from algebra that the graph 00:15:05.104 --> 00:15:18.801 of f of x equals 1 over x looks like this, roughly. It has a vertical asymptote at x equals 0, 00:15:18.803 --> 00:15:25.402 a horizontal asymptote at y equals 0. This graph is just going to be the same graph 00:15:25.404 --> 00:15:31.100 shifted over 2 to the right where the vertical asymptote would then be at x equals 2. 00:15:31.102 --> 00:15:45.501 So the graph of this function will look roughly like this, and what we're interested in 00:15:45.503 --> 00:15:52.106 is what happens as x approaches 2 from the left. So let's take a look. Well, 00:15:52.108 --> 00:15:58.401 as x approaches 2 from the left, the values of the function are going down and so 00:15:58.403 --> 00:16:04.904 approaching negative infinity. Now again, the definition of the limit says single 00:16:04.906 --> 00:16:10.604 real number, L. We're actually looking at the limit from the left this time, and so 00:16:10.606 --> 00:16:15.001 technically this limit does not exist, but again we're going to describe the behavior. 00:16:15.003 --> 00:16:22.900 We will describe this limit from the left, and we will use negative infinity. 00:16:22.902 --> 00:16:27.800 When possible, it's a good idea to sketch a graph when it's simple like this 00:16:27.802 --> 00:16:32.501 because you get a clear view of how you could read the limit, not only from the left. 00:16:32.503 --> 00:16:38.102 You could also read the limit from the right, which would be positive infinity, and then 00:16:38.104 --> 00:16:45.708 the limit as x approaches 2 would be does not exist. Ok, another example. Number 9. 00:16:45.710 --> 00:16:50.707 We're going to take the limit as x approaches 5 from the right. Again we have a function 00:16:50.709 --> 00:16:57.508 where direct substitution or simplification neither is going to be a possibility. So let's see if we can 00:16:57.510 --> 00:17:06.106 sketch the graph as we did on the previous example. Well, this function is a transformed 00:17:06.108 --> 00:17:14.302 version of 1 over x squared. So if you remember from algebra, 1 over x squared looks 00:17:14.304 --> 00:17:18.899 roughly like this. Notice that when you have negative values of x and you square them 00:17:18.901 --> 00:17:23.501 they're going to be positive. That's why the behavior of this function looks like this, 00:17:23.503 --> 00:17:29.302 and looks like this, with a vertical asymptote of x equals 0, horizontal at y equals 0. 00:17:29.304 --> 00:17:39.501 We take this graph, and we shift it 5 to the right. The behavior is not going to change, 00:17:39.503 --> 00:17:47.303 so we have a vertical asymptote at x equals 5, a horizontal asymptote at y equals 0, 00:17:47.305 --> 00:17:55.501 and again we have similar behavior. So we read the limit as x approaches 5 from the right, 00:17:55.503 --> 00:18:00.202 and as we come in from the right, the functional values are increasing, rising, 00:18:00.204 --> 00:18:05.701 approaching positive infinity. Again, the limit does not exist from the right, technically, 00:18:05.703 --> 00:18:12.212 but we describe the behavior with positive infinity. Just as a reminder on this function, 00:18:12.214 --> 00:18:30.203 at the limit as x approaches 5 from the left of the function 1 over x minus 5 squared. 00:18:30.205 --> 00:18:37.402 As we look at the limit from the left, we can see that as we come in to x equals 5 from the left, 00:18:37.404 --> 00:18:43.202 the functional values are increasing. The limit does not exist, but we describe the behavior 00:18:43.204 --> 00:18:51.696 again with positive infinity. And so, the limit as x approaches 5 from the right is described with 00:18:51.698 --> 00:18:57.804 positive infinity, and the limit as x approaches 5 from the left is described as positive infinity. 00:18:57.806 --> 00:19:04.202 But don't get too carried away and think that means that the limit as x approaches 5 is going 00:19:04.204 --> 00:19:16.501 to be positive infinity because it's not. Instead, this limit does not exist because these are 00:19:16.503 --> 00:19:22.103 not single real numbers. We just use this to describe the behavior. This is going to help us 00:19:22.105 --> 00:19:30.405 later when we start to graph. Let's take a look at an example where you can't do this. 00:19:30.408 --> 00:19:39.001 Example 10. We want to find the limit as x approaches 0 of the absolute value of x over x. 00:19:39.003 --> 00:19:47.002 Direct substitution will give us a zero in the denominator. Algebraic simplification is 00:19:47.004 --> 00:19:54.100 not a possibility here, and we may not be able to recall the graph immediately. 00:19:54.102 --> 00:19:59.303 So we're going to go to another possibility, and that is to make a table of values. 00:19:59.305 --> 00:20:14.504 So let's make a table. Let's pick values to the left of 0 and then values to the right of 0. 00:20:14.506 --> 00:20:24.600 We'll start with x equals negative 2. That gives us the absolute value of negative 2 over negative 2, 00:20:24.602 --> 00:20:35.200 so this is going to be 2 over negative 2, which is negative 1. Let's pick a value of x closer to 0. 00:20:35.202 --> 00:20:41.500 So let's try negative 1. Now you don't have to pick [integers]. They're just easier to work with. 00:20:41.502 --> 00:20:48.701 So when we do this, we get the absolute value of negative 1, all over negative 1, and so here 00:20:48.703 --> 00:20:56.101 I have 1 over negative 1, which is negative 1. So maybe we've established a pattern here. 00:20:56.103 --> 00:21:03.402 Let's take something even closer to 0, like negative 1/2. This would be the absolute value 00:21:03.404 --> 00:21:13.400 of negative 1/2, over negative 1/2. That's going to be positive 1/2 over negative 1/2, 00:21:13.402 --> 00:21:20.700 which is negative 1. At that point you realize that no matter what you sub in here 00:21:20.702 --> 00:21:26.800 as long as this number is negative, you're going to end of taking the absolute value of it here. 00:21:26.802 --> 00:21:33.397 So you will get positive whatever the number is, divided by negative of whatever the number is, 00:21:33.399 --> 00:21:42.702 and you're going to get negative 1. So what that means is, the limit as x approaches 0 00:21:42.704 --> 00:21:54.801 from the left of this function is negative 1. Now let's take a look at the limit from the right. 00:21:54.803 --> 00:22:04.803 I think we'll look at 2 and then possibly 1 and 1/2. We may see a pattern before that time. 00:22:04.805 --> 00:22:13.800 Let's put the absolute value of 2 over 2, and this will be 2 over 2 which is positive 1. 00:22:13.802 --> 00:22:23.501 If we use x equals 1, we get the absolute value of 1 over 1 which is 1. If we go further, 00:22:23.503 --> 00:22:27.702 you can see that what we're going to have here is the absolute value of a positive number 00:22:27.704 --> 00:22:34.003 is a positive number, over itself, and we're going to get 1. So what we have here is the 00:22:34.005 --> 00:22:47.403 limit as x approaches 0 from the right of this function is 1, and if you recall, 00:22:47.405 --> 00:22:58.700 we had the limit as x approaches 0 from the left of this function to be negative 1. 00:22:58.702 --> 00:23:00.000 And so what do we have? 00:23:00.002 --> 00:23:02.008 We have the limit from the left does not equal the limit from the right, 00:23:02.010 --> 00:23:10.402 and therefore, this limit does not exist. 00:23:10.404 --> 00:23:15.930 That concludes our examples on limits from equations. Let's review briefly 00:23:15.932 --> 00:23:20.934 some of the techniques we used. We started with polynomial functions, and 00:23:20.936 --> 00:23:26.135 we were able to use direct substitution in that case. Then we got to some functions 00:23:26.137 --> 00:23:32.432 where we could not use direct substitution until we algebraically simplified first. 00:23:32.434 --> 00:23:36.731 Another technique we used was to sketch the graph of the function, and then 00:23:36.733 --> 00:23:42.930 the final technique was to make a table of values. The decision as to 00:23:42.934 --> 00:23:49.299 which technique to use generally comes from an analysis of this function.