WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about continuity and one-sided limits.
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By now, we have a pretty good intuitive understanding of how limits work, and a reasonable sense of how we can evaluate them precisely.
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However, we haven't really worked with piecewise functions yet.
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In this lesson, we will not only learn how to apply limits to piecewise functions, but in doing so,
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we will learn about some new ideas involving limits.
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Our exploration will deepen our understanding of what it means for a function to be continuous, and also reveal the idea of a one-sided limit.
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And if you don't quite remember how piecewise functions work, or you find them confusing,
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make sure that you check out the lesson on piecewise functions that we did a long, long, long time ago,
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almost 40 lessons back--maybe even more; we did a lesson on piecewise functions.
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So, if you don't have an understanding of how piecewise functions work, you probably want to check that out first,
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because the idea of piecewise functions will end up showing up a lot when we are working with limits, especially in this lesson.
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All right, let's go: first we will set up a motivating example, so that we can have some realizations come out of this motivating example.
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This piecewise function example will help lead us to some ideas.
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We have -1 when x is less than 0; that is the portion right here.
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We have 0 when x equals 0 (that is the portion right there); and we have positive 1 when x is greater than 0.
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And notice how there are holes at actually equal to 0 on the top and actually equal to 0 on the bottom,
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because it jumps to the middle portion, right here, when it is actually at x = 0.
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OK, notice that, while f(0) exists (we get a value--we get f(0) = 0), the limit as x goes to 0 does not exist.
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As we show up from the two different sides, as we come in, they don't agree with each other.
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So, the point in the middle, f(0), does exist; but the limit does not exist.
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The issue is not so much that f(x) is a weird thing; it is not particularly weird.
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It is more an issue of f(x) all of a sudden doing this jump, where it is doing this giant leaping around.
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It is jumping from one place to another...wait a second...jumping?
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That sounds familiar--the idea of moving around suddenly...breaking the graph...
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Long, long ago, in this course, when we first learned about piecewise functions, about 40 lessons ago, we mentioned the idea of a continuous function.
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At the time, we lacked the formal ideas to precisely define continuity.
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So, we intuitively defined it as being any of these three equivalent things:
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all of the parts of the function are connected--we have some function, and everything in the function connects together;
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alternatively, we can think of this as that the function can be drawn without ever having to lift your pencil from the paper,
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because if you have to lift your pencil, if we draw part of it, but then you have to stop here,
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and lift your pencil to go somewhere else to keep going, well, we have this break here;
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and because of that break, it means that it is not continuous.
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And finally, there are no breaks or holes in the graph.
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We see a break pretty clearly in this, where we go one way, and then all of a sudden it switches to something else.
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But we can also it just having a hole as being the same issue, where it goes up here,
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but then for some reason this point is not defined, and then it keeps going.
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It is continuous, except for the part where it has this little break in the middle, which breaks continuity.
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At this point, we actually now have the formal background to expand on this intuitive definition of it not being connected,
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it not being continuous, and actually define continuity in a precise term, using limits.
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To help us see this, we will compare a piecewise example with a continuous function.
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Here is our piecewise example that we have been working with so far, and here is our continuous function friend
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for any time we need to pull out a random function, g(x) = x², a parabola.
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Over here in our piecewise break, we can interpret the break in continuity, this part where it jumps, as it jumping.
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The issue with it being continuous isn't that something breaks in the function.
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The function is perfectly defined at every number; any number that you put in...it knows what to go out to.
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But the issue is that it jumps all of a sudden; that is, at x = 0, it doesn't go where we expect; it does not go to the place that we expect.
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Either direction we come at this thing, no matter how we look at it, we can't build an expectation.
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We can't build an expectation for where the next place is, because there is no agreement; we can't agree on where we are going.
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Formally, when we can't have an expectation--there is no agreement on where we are going
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as we come from the two sides--that means that there is no limit as it goes to x = 0.
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As we get towards x = 0, limit as x goes to 0 does not exist, because there is no agreement on the value that they are going to.
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Those two sides are going to two totally different values.
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On the other hand, in g(x), there are no jumps; the nice thing about g(x), the nice thing about the parabola,
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is that the function goes where we expect; it always ends up showing up where we expect.
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Every point is exactly where we expect that point to be.
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That is to say, every point always has a limit, and that point matches with that limit.
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All of g(x) is exactly where we expect it to be.
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The limits and the points in the function are exactly the same thing.
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This is the idea that we use for continuity: we see that a break in continuity, the break in a function,
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is caused by the limit...our expectation not matching up with what the function does.
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Formally, we say that a function f(x) is continuous at some location c if f(c) exists and the limit as x goes to c of f(x) equals f(c).
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In other words, the function gives the value we expect when we evaluate it at c.
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The value that we expect to come out as we get close to c is exactly what the value ends up being.
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We expect some value, and then that value turns out to actually be what it is.
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That idea of the limit matching what the function actually is there is what it means for something to be continuous at a point.
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Expanding on this idea, we have more vocabulary--just ways to talk about this thing.
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If f(x) is not continuous at c, we say that it is **discontinuous** at c; the opposite of continuous is discontinuous.
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If we want to talk about some specific location where it breaks from being continuous, we say that it is a discontinuity.
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A single discontinuous point is a discontinuity.
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If f(x) is continuous at every point in an interval, over an entire interval--
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it is always continuous throughout that interval--we say f is continuous on (a,b), on some interval.
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If f(x) is continuous for every real number--it is always continuous--we simply say that f is a continuous function.
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The function is continuous if it always ends up working out continuously; great.
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Notice that the vast majority of the functions that we are used to working with are continuous functions.
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When we think of functions, we usually think of these nice, smooth, curved things that all fit together quite cleanly.
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So, when we think "normal," what we are really thinking is "continuous."
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This is why normal functions allow us to simply plug in the value we are approaching--
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because we generally have...a "normal" function, normally, in our minds, means that it is a continuous function.
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And by definition, a continuous function is one where the limit as x goes to c of f(x) is the same thing as f at c--
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that we can just plug in the c for our x, and we will end up getting the same thing as if we looked at the limit.
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The expectation is where we actually come out to be.
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Furthermore, this idea also explains why a function that is weird in one place still allows us to plug in the value we are approaching if it is not the weird value.
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That is because, if it is not the weird value, then that means that it is normal in that area;
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or more accurately, it is continuous in that place, if it is not the weird place.
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And because it is continuous in that place, that means we still have the same thing: that the limit as x goes to c of f(x) will match with f(c).
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That means that we can find out what the limit is by simply plugging in c into our function.
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So, the same logic is what we can apply here.
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And we end up seeing that, as long as it is normal in this region around c, as long as there is
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some tiny little neighborhood around c that behaves normally (which is to say continuously),
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we can just plug c in, and we will end up being able to figure out, "Oh, that is what the limit comes out to be."
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Let us return to our motivating example, now that we have the notion of continuity learned.
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Back to our motivating example: we have -1 when x is less than 0, 0 when x equals 0, and 1 when x is greater than 0.
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On the one hand, as we approach 0, we totally don't have a limit.
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As x goes to 0, it does not produce a limit, because they don't agree.
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If they don't agree, it is not a limit; both sides have to agree.
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However, if we were to imagine only approaching from a single side, all of a sudden, it starts to work; we could find limits.
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This realization leads us to the creation of a new type of limit, the one-sided limit.
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If we approach only from the negative side, it is clear that we are headed to -1.
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If we were to approach only from the positive side, it is clear that we would be headed to positive 1.
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So, we can have this idea of a one-sided limit, where we just look at approaching from one side or the other.
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We don't worry about what the other side is doing; we just worry about what this side is doing;
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where is it headed towards, as we stay coming from this one direction?
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A one-sided limit basically works just the same as a normal limit, except that it only considers one side; it is only looking at one side.
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The limit as x goes to c with a negative symbol in the top right--our c that we are going to, and then a negative symbol in the top right--
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denotes the limit as x approaches c from the left, or the negative side, equivalently.
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It is the limit as x goes to c, while x is less than c.
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So, as x remains less than c--that is, as we come from the left side, the negative side, that is the limit that we end up getting:
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limit as x goes to c, with that little negative sign in the corner.
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Similarly, if we have limit as x goes to c with a positive sign in the corner, that denotes the limit as x approaches c from the right, or positive, side.
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As we come from the right side, the positive side--that is going to be all of the x's where x is greater than c.
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So, as our x stays above c, what are we going towards?
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Be careful to keep track of which symbol goes with which side.
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They are easy to get confused at first, so be careful about this.
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Remember: the negative symbol means to look at the negative side; and positive means to look at the positive side.
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If we say that x goes to c with a negative symbol in that top right corner,
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that means that we are saying to look as we come from the left side, the negative side.
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If it is x goes to c with a positive sign in that top right corner, we are saying,
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"Look at what happens as we come from the right side, as we come from the positive side."
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All right, let's see this idea applied to our example.
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We have -1 when x is less than 0; so as we go in from the left side, limit as x goes to 0 with a negative sign,
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we get -1, because that is the value that we appear to be approaching.
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If we come from the positive side, the limit as x goes to 0 from the positive side, we end up approaching positive 1.
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If we just simply evaluate what f at 0 is, we get just simply 0.
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Now, notice: there is not necessarily any connection between the left and right side or the actual value of what the function is.
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-1 is totally different than 0, which is totally different than positive 1.
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The limits, the two side limits, don't agree, and the actual value that comes out of the function doesn't agree with either of them.
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So, each thing can behave totally independently.
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However, one connection is that the two different side limits don't agree.
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And because they don't agree, that means that there can't be a normal limit, the limit of it coming from both sides.
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For there to be a normal limit, they have to match up; if they don't match up, then that means that there is no normal limit.
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So, matching up: if the left and right side, the two sides, match up together, that means that we do have a normal limit.
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If, on the other hand, they don't match up, then that means that we don't have a normal limit.
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So, normal limits and one-sided limits: if both sides of a function go to the same value
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as some location is approached, then the normal limit exists there.
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Furthermore, if a normal limit exists, then both sides must go to the same value as they approach that location.
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Symbolically, we can write this as the left-side limit equals l, and the right-side limit equals l,
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implies that the normal limit is equal to l, as well.
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If our left side goes to the same value as our right side, then that means that the sides match up.
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So, we have a normal limit; and because they both went to l, then it comes out as an l limit, a limit going to l.
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Similarly, if we went the other direction, if we know that there is a limit here at l;
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then that means that, if we went off to the left side, or we went off to the right side,
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we are going to end up having it be l in both of those directions, as well.
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So, if the normal limit exists, then we know that the left-side limit and the right-side limit,
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the negative limit and the positive limit, have to both be equal to the same thing.
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They have to be equal to this same l.
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All right, continuing with the same idea: if both one-sided limits do not go to the same value--
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that is, they don't match up, or one of them simply doesn't exist at all--then a normal limit does not exist there.
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That is to say, if the left-side limit is not the same as the right-side limit, then that means that the normal limit does not exist.
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And if the normal limit does not exist, that tells us that one of them has to not exist, or that the sides don't match up to each other.
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So, there is this connection between normal limits and one-sided limits.
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If the one-sided limits meet up together, there is a normal limit at that place.
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If there is a normal limit at that place, then that means that the one-sided limits must match up together.
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And the same thing happens if the one-sided limits don't match up: then that means that there is no normal limit.
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If there is no normal limit, then we can't have matching up; great.
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All right, limits of piecewise functions: we are now ready to actually talk about how to apply all of this stuff to a piecewise function.
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It is this idea about "Do they match up? Do they not match up?" that lets us find the limits of a piecewise function at a break-over point.
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A break-over point is switching from one function to another function.
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We are going here, and then, all of a sudden, we come out as some other thing.
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Let's see what we are talking about here: it is some function here, and then all of a sudden, it swaps over to being some other function, like this.
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So, I am talking about this break-over place being where it switches between the two.
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If it is a break-over point, and we want to evaluate this, well, we can say, "Where were you going from the left side?
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Where were you going from the right side?" and we can ask questions about each of the one-sided limits and how they relate to each other.
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And if it is not a break-over point, then we don't have to worry about that, because that normally means
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that we are just in the middle of a normal, nice, continuous part of a piecewise function.
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And that means that we can usually just plug the location, as we discussed in the previous lesson, and as we sort of discussed earlier in this lesson.
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If you are in the middle of a nice continuous chunk of a function, you just plug in the location that you are going to,
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because "continuous" means that the limit matches up with what it evaluates to be there.
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So, if we do have a piecewise function, though, and we are looking at a break-over,
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where we switch from one track to another track, then we can find the limit of a piecewise function
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by checking if left and right side agree; if the left side and the right side agree with each other,
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then that means that we have a limit there; so if they agree, the limit exists.
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And it is equal to what they both are, because they have to match up to the same value.
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If they do not agree (one goes here and the other one goes here), then there is a jump between them,
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which means that there can't be a normal limit, because we have to agree from both sides; then, the limit does not exist.
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Since piecewise functions are usually made up of fairly normal functions
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(you don't really normally see any very weird functions when we are dealing with piecewise functions),
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since each side is normally fairly normal, it is normally pretty easy to find the left and right side limits,
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when we want to compare them, because we can just say, "OK, if we plug in for the left side over here,
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and plug in for the right side over here, we are able to figure out that that is what the left-side limit is;
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that is what the right-side limit has to be; now decide whether or not they actually match up."
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So, you can normally pretty easily figure out what they are.
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All right, let's see some examples: the first example: Find both of the one-sided limits below
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for f(x) = -3x + 3 when x < 2, and x² + 2 when x ≥ 2.
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First, let's just see a quick sketch to get some idea of how this thing works.
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-3x + 3 when x is less than 2: we are going to end up being at a fairly steep incline like this.
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And it pops out of existence here, because now we hit x < 2.
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And then, we switch over to some x² + 2; x is greater than or equal to 2;
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so, at x² + 2, we would pop up here; and then we would continue on our way,
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with whatever the parabola is, already in motion.
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All right, that is what we are seeing from this piecewise function.
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When it says "the limits below for x going to 2 from the negative side," well, if it is from the negative side,
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we are looking for what is happening as we come in here.
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So, if we are looking at what is coming in here, then that is x less than 2, which means we are just concerned about -3x + 3.
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So, if that is the case, then that means limit as x goes to 2 from the left side is just...
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if we plugged in -3x + 3, and we plug in 2 (it is 2 from the left side; we plug in 2), -3(2) + 3 is -6 + 3, so we get -3 there.
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Over here, if we want to talk about the limit as x goes to 2 from the positive side,
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then what we are concerned about is as we come in from this side; as we come in from this side,
00:18:20.800 --> 00:18:26.000
we belong to x being greater than or equal to 2; we don't actually have to worry about the x = 2,
00:18:26.000 --> 00:18:29.700
because they are both limits, so it is only the journey towards that location.
00:18:29.700 --> 00:18:33.400
We use x² + 2 if we want to talk about what it is going to be.
00:18:33.400 --> 00:18:37.400
x² + 2 was nice and normal until that flip-over; so we can just plug into that.
00:18:37.400 --> 00:18:47.000
x² + 2 means it is going to be 2² + 2, 4 + 2, or 6.
00:18:47.000 --> 00:18:50.300
And that is what we get for our two different sides for the limits here.
00:18:50.300 --> 00:18:52.300
Notice that they don't match up.
00:18:52.300 --> 00:18:57.400
All right, the second example: Evaluate the limit as x goes to 3 from the positive side of √(x - 3) + 4.
00:18:57.400 --> 00:19:01.800
First, let's just quickly draw a quick sketch, so that we can see what is going on here.
00:19:01.800 --> 00:19:07.900
So, √(x - 3) + 4: where would that start?
00:19:07.900 --> 00:19:12.600
Well, the first value that would make sense in there...if we plug in anything less than 3,
00:19:12.600 --> 00:19:15.600
we are going to be taking the square root of a negative number, so it can't be any less than 3.
00:19:15.600 --> 00:19:19.900
So, the very first point we could plot is at +3; and we would have 4 come out of it.
00:19:19.900 --> 00:19:23.500
And then, it would curve out like a square root function, like this.
00:19:23.500 --> 00:19:30.000
OK, if we are concerned about x going to 3 from the positive side, then what we care about is what happens on our way there.
00:19:30.000 --> 00:19:34.400
On our way there, it ends up just behaving like √(x - 3) + 4.
00:19:34.400 --> 00:19:39.900
It is totally normal, up until the moment where it stops existing completely, when we try to take square roots of negatives.
00:19:39.900 --> 00:19:48.900
The limit as x goes to 3 from the positive side, of √(x - 3) + 4...
00:19:48.900 --> 00:19:53.500
well, since it behaves totally normally up until the point where it stops existing,
00:19:53.500 --> 00:20:01.800
but we don't care about that side, because we are on the existing side, the positive side--we can just plug in our value for 3.
00:20:01.800 --> 00:20:11.200
√(3 - 3) + 4 is √0 + 4, or positive 4; and that totally makes sense.
00:20:11.200 --> 00:20:15.800
The point that we expect to go to is this one right here, where it starts.
00:20:15.800 --> 00:20:19.300
There are no weird jumps; there is nothing weird going on as we go in.
00:20:19.300 --> 00:20:30.300
As we come in from the positive side, it is very clearly headed towards 4, so it makes total sense that it has a limit.
00:20:30.300 --> 00:20:36.400
Now, what about explaining why the limit as x goes to 3 of √(x - 3) + 4 does not exist?
00:20:36.400 --> 00:20:43.000
Well, the real thing here is: does it come into the same thing from both sides?--no, because there is nothing over here at all.
00:20:43.000 --> 00:20:45.500
If we try to come in from the left side, what is going on?
00:20:45.500 --> 00:20:48.300
We have no idea what is going on; that is why it does not exist.
00:20:48.300 --> 00:20:56.500
If we want to get even more formal, we can say that, for the limit as x goes to 3 to exist,
00:20:56.500 --> 00:21:07.300
that must mean that the limit as x goes to 3 from the positive side is the same thing as limit as x goes to 3 from the negative side.
00:21:07.300 --> 00:21:19.400
But clearly, the limit as x goes to 3 from the negative side of f(x) (this being f(x) in this case) does not exist.
00:21:19.400 --> 00:21:24.100
You can't say that it is something that we are headed towards, because it just doesn't exist in that area.
00:21:24.100 --> 00:21:31.100
If it doesn't exist, as we try to come in from the left side, there is no limit to say that it is going to.
00:21:31.100 --> 00:21:38.200
So, if the right side simply does not exist, then that means that the normal limit can't exist, because we only have one-half of a normal limit.
00:21:38.200 --> 00:21:48.800
Therefore, the limit as x goes to 3 does not exist; and that is why we see that.
00:21:48.800 --> 00:21:53.200
But more informally, it is just the fact that we can't see it coming to the same thing from both sides.
00:21:53.200 --> 00:21:59.500
We have to have it from both sides; so if one side simply just isn't there, then it doesn't exist.
00:21:59.500 --> 00:22:04.100
The third example: Determine if the limit exists; if so, evaluate it.
00:22:04.100 --> 00:22:16.300
We have a piecewise function here: x³ + 4 for x < -1; 7 when x = -1; and (x - 1)² - 1 when x > -1.
00:22:16.300 --> 00:22:18.600
So, our first question is, "Where are we headed towards?"
00:22:18.600 --> 00:22:25.600
We are headed to x going to -1; so we don't just have it being in a nice, convenient, normal section.
00:22:25.600 --> 00:22:32.100
We are going exactly in the break-over; all right, well, if that is the case, how do we check to make sure it exists?
00:22:32.100 --> 00:22:38.100
Well, it only exists if from the right side and the left side it meets up to the same value.
00:22:38.100 --> 00:22:42.400
The two different sided limits have to agree with each other.
00:22:42.400 --> 00:22:54.600
If that is the case, what we are looking for is the limit as x goes to -1 of f(x); it is going to be based on the limit,
00:22:54.600 --> 00:23:00.700
as x goes to -1 from the negative side (and don't get confused about the -1 and then the negative;
00:23:00.700 --> 00:23:05.100
the negative in the top right tells us that we are coming from the negative side; the negative in the normal place
00:23:05.100 --> 00:23:14.600
just means that it is a negative number), is going to be the same thing as the limit as x goes to -1 from the positive side.
00:23:14.600 --> 00:23:22.000
If they end up being the same value for the limit, then we end up getting that it does exist as an actual limit, and it is that location.
00:23:22.000 --> 00:23:27.900
The first thing to notice at this point is...do we care about 7 when x = -1?
00:23:27.900 --> 00:23:31.600
No, we don't care about it, because it is x going to -1.
00:23:31.600 --> 00:23:35.100
And remember: it is about the journey, not the destination; it is about where you are headed,
00:23:35.100 --> 00:23:37.800
but you don't actually care about where you show up, when you are looking at a limit.
00:23:37.800 --> 00:23:40.300
You just care about where it seemed like you were going to.
00:23:40.300 --> 00:23:45.200
So, where we actually end up going doesn't matter; we don't have to worry about 7 at all.
00:23:45.200 --> 00:23:47.300
It is just there as a distraction to get us confused.
00:23:47.300 --> 00:23:53.300
The only things that we really have to care about are the x³ + 4, the part that we are coming from on the left side,
00:23:53.300 --> 00:24:02.200
x less than -1, and the (x - 1)² - 1, the part that we are coming from on the positive side, because it is x greater than -1.
00:24:02.200 --> 00:24:12.000
So, if that is the case, let's work this out; we know that our left side limit will be based off of x³ + 4,
00:24:12.000 --> 00:24:23.700
so that the limit as x goes to -1 from the negative side of f(x) will be the same as if we had simply plugged in -1 for our x here,
00:24:23.700 --> 00:24:31.800
since we are coming from the left side, and x³ + 4 is the way that the thing behaves as long as it is on the left side of -1.
00:24:31.800 --> 00:24:43.400
So, we plug into there; we have (-1)³ + 4; -1 cubed is just -1, plus 4...so we get +3 from the left side.
00:24:43.400 --> 00:24:48.800
Switch to looking at our right side now; the question is, "Do the two sides agree, or do they disagree?"
00:24:48.800 --> 00:24:55.600
If they go to different places, then the limit does not exist; if they go to the same place, then they do exist, and the limit is where they meet up.
00:24:55.600 --> 00:25:02.100
So, in this case, we are now looking at...the right-side function is (x - 1)² - 1.
00:25:02.100 --> 00:25:10.200
We are looking at the limit, as x comes in from the right side (that is, the positive side), of f(x).
00:25:10.200 --> 00:25:16.100
Once again, all we are concerned with is the right side of this, and the right side is entirely determined by this function,
00:25:16.100 --> 00:25:21.600
because how that is how the right side is behaving that whole time, just as the left side was behaving the whole time for x³ + 4.
00:25:21.600 --> 00:25:27.500
So, it is just a question of how that will fit into there, because (x - 1)² - 1 is a nice continuous function.
00:25:27.500 --> 00:25:31.300
So, we can just plug into it, because we don't have to worry about anything weird happening.
00:25:31.300 --> 00:25:40.800
We plug into that, and we have...-1 swaps out for our x, minus 1, squared, minus 1; -2 inside of there, squared, minus 1;
00:25:40.800 --> 00:25:47.700
so that gets us 4 - 1, which gets us 3; look at that: 3 and 3 check out.
00:25:47.700 --> 00:25:50.700
So, that means that indeed the limit does exist.
00:25:50.700 --> 00:25:56.500
Thus, we can combine these two things, and we know that, since the left- and right-side limits agreed with each other--
00:25:56.500 --> 00:26:06.700
they both came out to be 3--that means that the limit as x approaches -1 from both sides of f(x) is equal to 3.
00:26:06.700 --> 00:26:10.400
I do want to point out to you, though, that f(x) is not continuous.
00:26:10.400 --> 00:26:18.300
Why? Because at x = -1, it jumps; so the left side expects to go to 3; the right side expects to go to 3.
00:26:18.300 --> 00:26:20.500
But when we actually get to 3, it jumps up to 7.
00:26:20.500 --> 00:26:23.100
So, because it jumps to somewhere else, it is not continuous.
00:26:23.100 --> 00:26:25.700
The limit exists, but it is not actually a continuous function.
00:26:25.700 --> 00:26:31.900
It has to have a limit and agree with what that point actually comes out to be, to be continuous.
00:26:31.900 --> 00:26:40.300
All right, our final one, where we do actually talk about continuity: Determine the value of a that makes f(x) continuous.
00:26:40.300 --> 00:26:45.100
What does it mean to be continuous? (I will write that out as cts, just because I am lazy).
00:26:45.100 --> 00:26:52.200
To be continuous: that means that the limit, as x goes to c, of f(x), is equal to f(c).
00:26:52.200 --> 00:26:58.600
Remember: we talked about this as the expectation for the function being the same thing as what we actually get out of the function.
00:26:58.600 --> 00:27:00.900
That is what it means for something to be continuous.
00:27:00.900 --> 00:27:05.300
The expectation, the limit, is the same thing as what we actually get out, f(c).
00:27:05.300 --> 00:27:09.600
So, to be continuous, the limit as x goes to c of f(x) must be equal to f(c).
00:27:09.600 --> 00:27:15.200
However, in this case, we don't just have to worry about limits; we also have to worry about the fact that this is a piecewise function.
00:27:15.200 --> 00:27:21.300
Since it is a piecewise function, we need to make sure that both of the pieces end up agreeing.
00:27:21.300 --> 00:27:24.900
The first question is, "Does the limit exist?"
00:27:24.900 --> 00:27:38.400
Well, our question here--the function will be continuous if the limit as x goes to 1 of f(x) is equal to f at 1.
00:27:38.400 --> 00:27:40.700
All right, that is the case; so what is f at 1?
00:27:40.700 --> 00:27:43.100
We will come back to the left side and right side in just a second.
00:27:43.100 --> 00:27:48.400
So, what is f at 1? Well, f at 1 is equal to 5 minus...
00:27:48.400 --> 00:27:54.200
Oh, which one do we have to use? We use x less than or equal to 1, so we are using this right here.
00:27:54.200 --> 00:28:07.200
So, 5 - 2: we swap out the x for a 1, minus...swap out the x for a 1...squared; 5 - 2 - 1 comes out to be 2.
00:28:07.200 --> 00:28:12.800
So, f(1) comes out to be 2; great.
00:28:12.800 --> 00:28:17.800
All right, now what we need to figure out is if the limit exists there.
00:28:17.800 --> 00:28:21.700
If our limit is going to exist, we have two different sides that we are coming from.
00:28:21.700 --> 00:28:26.400
We are coming from the left side, and we are coming from the right side.
00:28:26.400 --> 00:28:29.000
We are coming from the less than or equal and the greater than.
00:28:29.000 --> 00:28:32.800
We have both a left and a right side that we have to make sure match up.
00:28:32.800 --> 00:28:44.100
Since we have two different possibilities, we have to make sure that the limit as x goes to 1 of f(x) is the same thing as saying
00:28:44.100 --> 00:28:52.700
(because we have to check that both sides match) that the limit, as x goes to 1 from the negative side,
00:28:52.700 --> 00:29:01.100
is equal to the limit as x goes to 1 from the positive side.
00:29:01.100 --> 00:29:07.000
Now, we have to make sure that they are the same; and they should, of course, both be f(x) in here.
00:29:07.000 --> 00:29:08.800
We have been talking about f(x) this whole time.
00:29:08.800 --> 00:29:13.000
The limit as x goes to 1 from the negative side: well, actually, we already figured it out.
00:29:13.000 --> 00:29:15.900
What is the limit as x goes to 1 from the negative side?
00:29:15.900 --> 00:29:20.700
Well, that is going to be based off of how this works, because it is the left side.
00:29:20.700 --> 00:29:29.500
x ≤ 1 behaves like 5 - 2x - x²; so if you are at the less than, you are on the negative side, so it behaves just like 5 - 2x - x².
00:29:29.500 --> 00:29:34.300
So, we already figured out what happens there; that comes out to be 2.
00:29:34.300 --> 00:29:38.900
Since we are behaving just like the left side, that is just like it is going towards 2.
00:29:38.900 --> 00:29:42.700
So, we know that this is going to come out to be 2.
00:29:42.700 --> 00:29:51.300
Really, our only question is, "Does this part here equal the limit as x goes to 1 from the positive side?"
00:29:51.300 --> 00:29:56.000
We are allowed to determine ax - 1; we can't change x, because that is just the variable;
00:29:56.000 --> 00:30:02.100
but a--we are supposed to determine the value of a that will make this whole thing continuous.
00:30:02.100 --> 00:30:11.100
We know 2, because that is what the limit is as x goes to 1 from the negative side, has to be equal to ax - 1.
00:30:11.100 --> 00:30:18.500
What x are we going to? We are going to 1 from the positive side, so we just use ax - 1; we plug in 1 for our x; minus 1.
00:30:18.500 --> 00:30:26.500
Start working that out: we have 2 = a - 1; we add 1 to both sides; we have 3 = a, so 3 must be equal to a.
00:30:26.500 --> 00:30:33.300
And if 3 is equal to a, then that means that the limit on the right side is equal to the limit on the left side,
00:30:33.300 --> 00:30:38.500
which means that the limit does exist, and that the limit will come out to be 2.
00:30:38.500 --> 00:30:45.100
And since the limit comes out to be 2, and we know that f(1) = 2, we now see that, yes, it is, indeed, continuous.
00:30:45.100 --> 00:30:49.300
It might be a little bit hard to understand what is going on, so it can really help to see this graphically.
00:30:49.300 --> 00:30:52.200
Let's draw a quick picture, just to cement our understanding.
00:30:52.200 --> 00:30:58.200
That is how you do it technically; but it is really useful to understand what is going on intuitively, as well.
00:30:58.200 --> 00:31:06.700
Our 5 - 2x - x² would graph basically like this.
00:31:06.700 --> 00:31:11.300
And when we get to x ≤ 1, we jump over to the other one.
00:31:11.300 --> 00:31:21.500
Now, since it is ax - 1, a is taking the position of the slope: mx + b is how we normally graph a line.
00:31:21.500 --> 00:31:28.700
So, mx - 1 means that we are shifted down 1; we are definitely going to have a point there, down one on the y-axis.
00:31:28.700 --> 00:31:32.300
But our a--we are allowed to choose our value of a.
00:31:32.300 --> 00:31:35.600
So, what we are doing is effectively choosing the slope that we are going to have.
00:31:35.600 --> 00:31:39.000
So, that means that we get to choose some possibility for our slope.
00:31:39.000 --> 00:31:50.500
We have any possible rotation of this line; all of the different lines that could go through here with various different slopes are all of the different possibilities.
00:31:50.500 --> 00:31:54.900
The one that we have to choose to make this thing come out to be continuous
00:31:54.900 --> 00:32:01.700
is where that line matches up to where we have this hand-off, where we have this break-over point.
00:32:01.700 --> 00:32:06.400
So, we choose the one that matches up; we choose that slope, and it continues out from here.
00:32:06.400 --> 00:32:14.800
And that way, we end up having that the break-over ends up changing to a new track, but that track starts in the same place.
00:32:14.800 --> 00:32:21.200
We have chosen the slope so that the line matches up with where the other one finished off; and it goes through; great.
00:32:21.200 --> 00:32:26.000
All right, that finishes this; we now have an understanding of how to deal with piecewise functions.
00:32:26.000 --> 00:32:32.200
It is basically a function of if the left side and right side, the one-sided limits, match up to each other.
00:32:32.200 --> 00:32:35.900
And if we are talking about continuity, it is a question of if the limit matches what comes out of it.
00:32:35.900 --> 00:32:39.700
And sometimes, it will become a question of if the two sides match what comes out of it.
00:32:39.700 --> 00:32:43.000
All right, that finishes this lesson; we will see you at Educator.com later--goodbye!