WEBVTT 00:00:01.503 --> 00:00:06.103 Now that you understand limits, let's take a look at topic 3, continuity. 00:00:06.105 --> 00:00:13.102 A function f is continuous at the point x equals c if all of the following 00:00:13.104 --> 00:00:19.002 three conditions hold. First condition: f of c must exist. 00:00:19.004 --> 00:00:21.802 That means the function must be defined at the point. 00:00:21.804 --> 00:00:29.304 Second condition: The limit as x approaches c of f of x must exist. That means that the 00:00:29.306 --> 00:00:35.901 limit as x approaches c from the left has to equal the limit x approaches c from the right. 00:00:35.903 --> 00:00:41.202 And then the third condition is that the limit as x approaches c of f of x, which is the 00:00:41.204 --> 00:00:48.501 value from the second condition, must equal f of c, which is the value from the first condition. 00:00:48.503 --> 00:00:52.705 So let's take a look at what that means. That means first of all 00:00:52.707 --> 00:00:57.600 that the function has to be defined at the point, so there's no hole or asymptote or break. 00:00:57.602 --> 00:01:00.005 The function is defined at the point. 00:01:00.007 --> 00:01:05.302 Second of all, that means the limit from the left and the right are defined, come in together, 00:01:05.304 --> 00:01:10.602 and all three of those numbers have the same value. That gives you a smooth, continuous flow, 00:01:10.604 --> 00:01:17.003 hence the name continuity, with no breaks or jumps. Let's take a look at some examples. 00:01:17.005 --> 00:01:23.695 Example 1. Let's take a look at the graph of this function. Usually the directions to 00:01:23.697 --> 00:01:28.300 something of this nature will say discuss the graph of the function. 00:01:28.302 --> 00:01:34.301 Well the first thing you notice is there's obviously a break here at x equals 2. 00:01:34.303 --> 00:01:44.101 Let's explain why, using the definition. The first condition says f of c must exist. 00:01:44.103 --> 00:01:52.901 We're discussing the area here at x equals 2, so I'm going to write f of 2, 00:01:52.903 --> 00:02:02.301 and this does exist, and it has the value 3. So the first condition is met. 00:02:02.303 --> 00:02:09.000 Second condition. This says the limit must exist, the limit as x approaches 2 00:02:09.002 --> 00:02:13.401 from the left and the right, and this is where we're going to see a problem here. 00:02:13.403 --> 00:02:29.599 The limit as x approaches 2 from the left of f of x, this function, is 1. 00:02:29.601 --> 00:02:44.601 The limit as x approaches 2 from the right of this function is 3, and so clearly the 00:02:44.603 --> 00:02:52.800 limit from the left does not equal the limit from the right. So the limit as x approaches 2 00:02:52.802 --> 00:03:03.300 of f of x does not exist. So what we have shown here is, what we can see by looking at 00:03:03.302 --> 00:03:15.100 the graph, that this function is discontinuous at x equals 2. So we write f is discontinuous, 00:03:15.102 --> 00:03:30.200 or not continuous, at x equals 2. Notice that this function is continuous everywhere else, 00:03:30.202 --> 00:03:39.400 except at x equals 2. This leads us to another statement that says a function is 00:03:39.402 --> 00:03:45.500 continuous on an open interval from a to b if it's continuous at each point on the interval. 00:03:45.502 --> 00:03:54.901 So this function is continuous up to 2 and then from 2 to infinity. Let's look at example 2. 00:03:54.903 --> 00:04:02.199 Here we have a situation where the function is undefined at x equals 2, 00:04:02.201 --> 00:04:06.700 so when we discuss the continuity of the graph, that's where we're going to focus. 00:04:06.702 --> 00:04:16.299 Let's take a look at the conditions. First condition, f of 2. Well, I guess I've said this 00:04:16.301 --> 00:04:22.801 function is undefined at 2, so we immediately see that f of 2 does not exist. 00:04:22.803 --> 00:04:28.301 Now since all three conditions have to hold, then we can stop at this point, and 00:04:28.303 --> 00:04:34.800 simply say the function is discontinuous at 2 because it doesn't meet the first condition. 00:04:34.802 --> 00:04:52.700 So we write, f is discontinuous at x equals 2 because it doesn't meet the first condition 00:04:52.702 --> 00:04:59.201 of the definition. Let's discuss the continuity of this graph in example 3. 00:04:59.203 --> 00:05:08.901 Looking at the definition again, we start with the first condition. It seems just by looking, 00:05:08.903 --> 00:05:17.001 that this is where the problem is going to be, again at 2, so let's first find f of 2. 00:05:17.003 --> 00:05:25.400 f of 2 is actually defined here at negative 1, so f of 2 exists. 00:05:25.402 --> 00:05:29.900 Second condition, let's look at the limit from the left and the right. 00:05:29.902 --> 00:05:44.701 The limit as x approaches 2 from the left for this function, f of x, would be 1, 00:05:44.703 --> 00:05:57.700 and the limit as x approaches 2 from the right of f of x is also 1. You can see at this point 00:05:57.702 --> 00:06:02.899 we can actually stop if we wanted to, as soon as we got to here and realize that the limit 00:06:02.901 --> 00:06:07.799 from the left was not going to match the value of the function here. But let's go ahead and 00:06:07.801 --> 00:06:17.501 complete the second condition. So that means the limit as x approaches 2 of f of x is 1, 00:06:17.503 --> 00:06:24.501 and so what happens is the third condition here is the one that is actually violated because the 00:06:24.503 --> 00:06:41.600 limit as x approaches 2 of f of x is 1, which does not equal f of 2, which is negative 1. 00:06:41.602 --> 00:06:52.000 So what do we have here? We have this side, all the way to here, this side 00:06:52.002 --> 00:06:57.601 does not equal this side. Condition three is not met, and that is why this function is 00:06:57.603 --> 00:07:18.000 clearly discontinuous at x equals 2. So we write f is discontinuous at x equals 2. 00:07:18.002 --> 00:07:23.901 Let's take a look at a fourth graph. You can see when you look at this that 00:07:23.903 --> 00:07:31.200 this curve is smooth, and you generally would use the word continuous. Let's explain why. 00:07:31.202 --> 00:07:38.099 Over the entire open interval from negative infinity to infinity there are no breaks, and 00:07:38.101 --> 00:07:43.801 at any point that you would select, for any value of x, you can see that the function exists 00:07:43.803 --> 00:07:48.998 because there are no holes or asymptotes. You can see that the limit from the left and the 00:07:49.000 --> 00:07:54.501 limit from the right would be the same at any point, and you can see that these values would 00:07:54.503 --> 00:08:00.400 all be the same. So therefore, when we discuss the continuity of this graph, we write 00:08:00.402 --> 00:08:15.100 f is continuous on, and we write the open interval from negative infinity to infinity. 00:08:15.102 --> 00:08:19.800 Now let's take a look at some examples where we don't have the graph. 00:08:19.802 --> 00:08:37.600 Example 5: f of x equals 3x minus 1. What we want to do is discuss the continuity of this function. 00:08:37.602 --> 00:08:42.827 Well, once again, we analyze the function, 3x minus 1. This is a linear function. 00:08:42.829 --> 00:08:47.828 It has a slope of 3, y-intercept of (0,-1). You can visualize the graph. 00:08:48.830 --> 00:08:53.525 It has a domain from negative infinity to infinity. It has no holes, no breaks, and so 00:08:53.527 --> 00:09:05.528 therefore, this function is continuous on its domain from negative infinity to infinity. 00:09:05.530 --> 00:09:09.627 This is a polynomial. Let's look at another polynomial function. 00:09:09.629 --> 00:09:20.927 Example 6: f of x equals x squared plus 3x plus 4. Again, we analyze the graph of the function. 00:09:20.929 --> 00:09:25.825 What do we see? Well, we have a quadratic function. The graph is a parabola. There's 00:09:25.827 --> 00:09:32.628 no restriction on the domain. Domain is all reals, negative infinity to infinity, no holes, no breaks, 00:09:32.630 --> 00:09:43.528 and so again, this function is continuous on the open interval from negative infinity to infinity. 00:09:43.530 --> 00:09:54.730 Alright, example 7. This is going to be f of x equals 1 over x minus 4, and now we have a 00:09:54.732 --> 00:10:00.327 rational function. Again, if you picture the graph of 1 over x minus 4, what you see is the 00:10:00.329 --> 00:10:10.127 graph of 1 over x, shifted 4 units to the right. What that means is that when x equals 4, 00:10:10.129 --> 00:10:17.024 you have a vertical asymptote. So you're going to have a vertical asymptote at x equals 4, and 00:10:17.026 --> 00:10:22.728 without actually drawing the graph, you can see that this is going to be a point of discontinuity. 00:10:22.730 --> 00:10:31.927 Let's look at why. First condition, f of c exists. Clearly, if you have a vertical asymptote, f of 4 00:10:31.929 --> 00:10:49.527 does not exist. The first condition is violated, and so we write f is discontinuous at x equals 4. 00:10:49.529 --> 00:10:58.027 Alright, let's look at some piecewise graphs. Example 8. This is a function defined piecewise. 00:10:58.029 --> 00:11:06.928 We have two pieces. We have 3x minus 1 and x plus 5. The domain here is broken at x equals 3. 00:11:06.930 --> 00:11:11.928 We're going to discuss the continuity of this function, and so the first thing we notice is that 00:11:11.930 --> 00:11:17.727 there is no equal sign under either one of these. That immediately tells us that this function, 00:11:17.729 --> 00:11:23.628 though it would be continuous over this piece and this piece, is not going to be continuous at 3 00:11:23.630 --> 00:11:35.227 because it violates the very first condition. f of 3 does not exist. Again, why? 00:11:35.229 --> 00:11:40.526 Because the function is not defined at 3. It's defined on either side of 3, so the limit from the left 00:11:40.528 --> 00:11:45.727 and the right would exist, but the value of the function at 3 does not exist, so 00:11:45.729 --> 00:11:59.728 f is discontinuous at x equals 3. Alright, let's look at another piecewise [function]. 00:11:59.730 --> 00:12:13.428 Example 9. This one is defined as f of x equals 3x minus 4 when x is less than negative 1. 00:12:13.430 --> 00:12:23.327 That's the first piece, and the second piece is x squared when x is greater than or equal to negative 1. 00:12:23.329 --> 00:12:29.700 Ok, the first thing that we notice about this is the equals sign. So this function is defined at 00:12:29.702 --> 00:12:39.999 negative 1 for this piece. Let's look at the value here. Condition one. f of negative 1. 00:12:40.001 --> 00:12:45.799 So we use this piece, we plug negative 1 into here. Negative 1 squared is positive 1. 00:12:45.801 --> 00:12:52.000 So the first condition is met. Second condition. The limit. We're going to look at 00:12:52.002 --> 00:12:59.500 the limit from the left and the limit from the right. Ok, the limit, that is at negative 1, 00:12:59.502 --> 00:13:07.300 so the limit as x approaches negative 1 from the left of this function, f of x, is ... 00:13:07.302 --> 00:13:12.399 Ok, now let's go back over how we're going to do this. Because this is a polynomial 00:13:12.401 --> 00:13:17.503 function, it's a line, it's going to be continuous everywhere that it is defined, which is 00:13:17.505 --> 00:13:22.100 from negative infinity to negative 1, so we can substitute in here to take the limit. 00:13:22.102 --> 00:13:28.000 So this would be negative 1 times 3, negative 3, minus 4 is negative 7. 00:13:28.002 --> 00:13:35.799 So the limit from the left as x approaches negative 1 exists. Alright, limit from the right. 00:13:35.801 --> 00:13:42.400 The limit as x approaches negative 1 from the right of f of x is... 00:13:42.402 --> 00:13:48.199 So which piece do we go to now? Well, these are values greater than negative 1, 00:13:48.201 --> 00:13:54.700 so we go to this piece. When we put negative 1 into here, again because this is a [polynomial] 00:13:54.702 --> 00:14:02.199 and we can substitute in, we get negative 1 squared, which is positive 1. And so what happens? 00:14:02.201 --> 00:14:06.300 The limit from the left does not equal the limit from the right, and therefore, the 00:14:06.302 --> 00:14:24.500 second condition is violated. The limit as x approaches negative 1 of f of x does not exist, 00:14:24.502 --> 00:14:30.898 and so therefore this function is discontinuous at x equals negative 1. 00:14:30.900 --> 00:14:41.899 Discontinuous at x equals negative 1. And why, again? 00:14:41.901 --> 00:14:51.400 Because the second condition of the defintion is violated. Intuitively, what that means is that 00:14:51.402 --> 00:14:59.100 this piece will not touch this piece because the limit from the left is going to the value negative 7. 00:14:59.102 --> 00:15:04.899 The limit from the right is going to the value 1, and obviously they're not going to touch. 00:15:04.901 --> 00:15:16.903 Example 10. Let's define this piecewise [function] as f of x equals, and we're going to have 00:15:16.905 --> 00:15:27.199 three pieces here. The first piece is 2x minus 1, and this is defined when x is less than 0. 00:15:27.201 --> 00:15:38.901 The second piece is 1. This is defined when x is equal to 0, so this is going to be the point (0, 1). 00:15:38.903 --> 00:15:47.999 And then the third piece is negative 1, and that will be when x is greater than 0. 00:15:48.001 --> 00:15:54.706 Now this is going to be a horizontal line from 0 to infinity. 00:15:54.708 --> 00:16:00.303 Let's take a look at the continuity of this function. If there is a problem, 00:16:00.305 --> 00:16:05.103 it is likely to exist at x equals 0 because this is where the break in the domains lies. 00:16:05.105 --> 00:16:11.904 Let's go through the conditions one at a time. First condition. What is f of 0? 00:16:11.906 --> 00:16:17.501 We go to the middle piece here, and we discover that when x equals 0, the functional 00:16:17.503 --> 00:16:24.303 value is 1. This is actually the point (0, 1). So the first condition is met. 00:16:24.305 --> 00:16:35.103 Second condition. Let's look at the limit as x approaches 0 from the left. 00:16:35.105 --> 00:16:41.703 Where would we go to find this? Well, we go to values of x less than 0. 00:16:41.705 --> 00:16:49.899 We substitute in 0 here, so we get 0 minus 1, which is negative 1. We can do that 00:16:49.901 --> 00:16:56.301 because this is a linear function. The graph is a line. It's defined for all numbers less than 0. 00:16:56.303 --> 00:17:04.902 The second part of this would be the limit as x approaches 0 from the right of f of x. 00:17:04.904 --> 00:17:09.400 So we go to the third piece. Now this one is a little different because there is no x here, 00:17:09.402 --> 00:17:16.300 so there's no direct substitution. What we have here is a horizontal line equal to negative 1, 00:17:16.302 --> 00:17:23.900 and so for any value greater than zero, the function has the value negative 1. 00:17:23.902 --> 00:17:35.903 So what do we have? Well, we get here that the limit as x approaches 0 of f of x is negative 1, 00:17:35.905 --> 00:17:43.399 and that meets condition two. But as you look ahead to condition three, you see that 00:17:43.401 --> 00:17:50.101 condition one has the value 1, condition two, negative 1. Those are not equal. 00:17:50.103 --> 00:18:13.403 1 does not equal negative 1. Condition three is violated, and so f is discontinuous at x equals 0. 00:18:13.405 --> 00:18:34.203 Let's take a look at one last example. f of x is defined to be x squared plus 1 when x is less than 2 00:18:34.205 --> 00:18:45.502 and x plus 3 when x is greater than or equal to 2. Again we want to discuss the continuity 00:18:45.504 --> 00:18:52.800 of this function. This is a quadratic function. The shape is a parabola. 00:18:52.802 --> 00:18:57.303 It will be defined continuously from negative infinity to 2. 00:18:57.305 --> 00:19:05.703 This is a line. It will be defined from, or at, 2 to infinity, and so the question is what happens at 2. 00:19:05.705 --> 00:19:12.501 So let's go through the three conditions. What is the value of the function at 2? 00:19:12.503 --> 00:19:17.400 We go to this second piece because this includes the value where x is equal to 2. 00:19:17.402 --> 00:19:25.999 When I sub in 2 here, I get 5. First condition met. Second condition. 00:19:26.001 --> 00:19:36.102 The limit as x approaches 2 from the left of f of x is, and we go to the first piece. 00:19:36.104 --> 00:19:44.102 We sub 2 into here. 2 squared is 4, plus 1 is 5, and we're looking good so far. We have a 5. 00:19:44.104 --> 00:19:52.304 Let's do the limit as x approaches 2 from the right of f of x. Well, we already know what's going 00:19:52.306 --> 00:19:58.504 to happen here because we actually did this same substitution here. This is the part where we 00:19:58.506 --> 00:20:04.599 approach 2 from the right, so we use these values. So we sub 2 in here, and we get 5. 00:20:04.601 --> 00:20:20.501 So that means the limit as x approaches 2 of f of x is 5, and the third condition. This 5 and this 5 00:20:20.503 --> 00:20:30.800 are equal, and so that meets condition three, along with condition two and condition one. 00:20:30.800 --> 00:20:38.700 That means that there is no break or discontinuity at x equals 2. Therefore, this function is 00:20:38.702 --> 00:20:49.201 continuous over the entire domain which goes from negative infinity to infinity, 00:20:49.203 --> 00:20:57.502 with no points of discontinuity. So what have we done here with continuity? 00:20:57.504 --> 00:21:05.702 We have used this three condition definition to explore the continuity of graphs 00:21:05.704 --> 00:21:11.304 and functions given in the form of equations. We go through this step-by-step, 00:21:11.306 --> 00:21:17.505 and as soon as we find one condition that does not hold, we have identified a point of discontinuity. 00:21:17.507 --> 00:21:23.600 We now have all the tools we need to develop the next major concept of calculus.