Continuity - Online Tutor, Practice Problems & Exam Prep

Continuity of a function at a point \( c \) occurs when the limit as \( x \) approaches \( c \) equals the function value at \( c \). Discontinuities arise from holes, jumps, or asymptotes, particularly in rational and piecewise functions. To identify discontinuities, analyze the function's limits and values. For rational functions, set the denominator to zero, while for piecewise functions, compare one-sided limits at transition points. Understanding these concepts is crucial for recognizing function behavior and ensuring accurate mathematical analysis.

Intro to Continuity

Hey, everyone. Here, we're going to talk about continuity. Now we've actually already been dealing with continuity throughout our last several lesson videos. So here, we're really just putting a name to it. See, a function is continuous at some point c if limrarrx=c=f(c). Now we've seen this happen for quite a few different functions, so we would say that these functions are continuous at those points. Now here, we're going to dive a bit deeper into continuity and also talk about what it means for a function to be discontinuous. So let's go ahead and get started.

Now let's first take a look at our function on the left here. If we want to find the limit of this function as x→2, I can see that as x gets really, really close to 2 from either side, my function is approaching a y-value of 4. Now if I look at the actual value of my function when x=2, I see that it's actually the exact same as my limit. It's also 4. So here, my limit is the same as my function value. Now because these two values are the same, this function is continuous at x=2.

But let's take a look at our other function here. Again, if I want to find the limit of this function as x→2, I can see that as x gets really, really close to 2 from either side, my function is approaching a y-value of 1. Now as for the actual value of my function when x=2, I see that there's actually a hole on my graph when x=2. So the value of my function here is actually undefined. So here, my limit is definitely not the same as my function value. So I know that this function is discontinuous for x=2.

Now by finding our limit and our function value, we're able to determine continuity using its definition. Now I also want to give you a sort of visual tool to recognize continuity. Now if we take a look at our graph and we take our pen or our pencil and we trace through our function, if we can trace through a specified point without stopping, then the function is continuous there. Now let me show you exactly what I mean by that. Coming back down to our parabola here, if I trace through my function when x=2, I can do so smoothly without having to lift my pen. So I know it's continuous there. But if I come over to my function here and I try to do the same thing, if I do that here I can start tracing, but then I have to stop, pick up my pen, and continue on the other side of x=2 because it is discontinuous there.

Now you can think of this as a sort of tool to help you identify problem areas on a function by looking at a graph, but you'll often still need to find your limit and your function value to fully verify continuity. Now with this function here, it was discontinuous at 2 specifically because there was a hole there. Now discontinuities do occur whenever a function has a hole, but also when a function has either a jump or an asymptote. Now discontinuities happen commonly for rational functions and for piecewise functions, with rational functions often having holes or asymptotes, and piecewise functions often having jumps, although they can have other types of discontinuities as well. Now with all of this in mind, let's go ahead and apply this knowledge to an example problem.

Now in this example, we're asked to determine if the function is continuous for each given value of c by finding both our function value and our limit. Now we know if these two values are the same that our function is continuous at that point, whereas if they are not the same, we know our function is discontinuous there. So let's go ahead and dive into this first value. Here we have c=-2. Now using my sort of visual tracing tool, I can take a look at my graph and see that if I trace along my function when x=-2, I don't have to pick up my pen. So I already have an idea that this function is continuous here. But here, we specifically want to verify continuity by using our limit and our function value. So let's go ahead and find that limit. Now as x gets really, really close to -2 from either side, I can see that my function is approaching a y value of 1. Now the value of my function here is actually also 1. So because the limit is the same as my function value, this function is continuous at -2. Well, let's move on to our next example here. Here we have c=4. So we first want to find the limit here as x→4. Now as x approaches 4 coming in from that left side, I can see that my function is approaching a value of 1 whereas coming in from the right side we're approaching a value of 3. So this limit actually does not exist because my function is not approaching the same value from both sides. Now as for our function value here, I can see that when x=4, my function value is 1. Now if my limit does not exist, I can sort of already see that it's going to be discontinuous here, and I can see further that these are not the same, does not exist is not the same as 1. So I know my function is discontinuous here at 4. And looking at this discontinuity specifically, I can see that this is a jump discontinuity because I start on one piece of my function, and I have to jump up to another. Now let's move on to our final example here. Here we have c=1. So looking at the limit of our function as x→1, I can see that from either side, my function is going down to negative infinity. But remember that whenever we have unbounded behavior, our limit does not exist. Now if we further look at our function value here, our function value is actually undefined because I can see that I have an asymptote here. So I know that this function is discontinuous,

Use graph of $f\left(x\right)$ to determine if the function is continuous or discontinuous at $x=c$

$c=0$

• A. Continuous
• B. Discontinuous

Use the graph of $f\left(x\right)$ to determine if the function is continuous or discontinuous at $x=c$.

$c=-2$

• A. Continuous
• B. Discontinuous

Use the graph of $f\left(x\right)$ to determine if the function is continuous or discontinuous at $x=c$.

$c=3$

• A. Continuous
• B. Discontinuous

Use the graph of $f\left(x\right)$ to determine if the function is continuous or discontinuous at $x=c$.

$c=4$

• A. Continuous
• B. Discontinuous

Determine Continuity Algebraically

We just learned how to determine the continuity of a function at a given point by using a graph, but often you won't actually be given the luxury of a graph. Instead of being given a specific point to test the continuity at, you'll be asked to specifically find where a function is discontinuous by just being given that function. Now remember that when we were working with graphs, we saw that discontinuities occurred whenever a function had a jump, a hole, or an asymptote, which happens often for rational functions and for piecewise functions. So that means that we need to be able to determine where these discontinuities are when working with both of these types of functions. Now here, I'm going to walk you through how to do that step by step. So let's go ahead and get started. Now let's first just jump right into our example here, and we want to determine the values of x here, if any, for which our function is discontinuous.

The first function that we're given here is f(x) = (x² - x - 6) / (x + 2). Polynomials by themselves are continuous everywhere, but because these polynomials are being divided and we're dealing with a rational function here, we know that this rational function may have holes or asymptotes. It would be discontinuous there. But how do we find where these discontinuities are? Well, for rational functions, it's actually rather easy to determine where these discontinuities are because it's just going to be where our denominator is equal to 0. So here our denominator is x + 2. If we take that denominator and set it equal to 0 and then just solve for x here by subtracting 2 from both sides, I end up with x = -2. This is where my function is discontinuous. One of 2 things could be happening here. This could be an asymptote, or if we were able to factor this numerator and cancel out one of the factors with the denominator, this would end up being a hole or a removable discontinuity.

As for a piecewise function, we want to do the same thing. We want to determine the values of x, if any, for which our function is discontinuous. Here our function is equal to 4 for x < -1, and it's equal to x + 1 for x ≥ -1. We don't need to worry about discontinuities within each piece, but because this is a piecewise function, there may be a discontinuity, like a hole or a jump, in between my two pieces. In order to determine if there is a discontinuity there, that means that we want to compare the limit and the function value for the point or points in between each piece. Here, since I only have 2 pieces, I only have to worry about one point here, and that's x = -1. For this point, I want to look at my limit and my function value. So I want to find the limit of f(x) as x approaches -1, and I also want to find my function value f(-1). If these values are equal to each other, I know that my function is continuous there. But if they are not equal to each other, I know that I have a discontinuity there.

Starting first with our limit, because this is a piecewise function, I need to be careful here and look at my one-sided limits. Coming in from the left side, my function is going to be equal to 4 for these values of x less than -1. So I know that that left sided limit is simply 4. For the right sided limit, the limit of f(x) as x approaches -1 from that right side. Looking back at my piecewise function, I know that coming in from that right side, my function is going to be equal to x + 1 because these values are greater than -1. So that means this limit is going to be equal to -1 + 1, which is simply 0. Now looking at my one-sided limits, one of them is 4 and the other one is 0. Because these are not the same, it tells me that my regular limit does not exist. I know that it's not going to be the same as my function value. But let's go ahead and continue on here. Now my function value f(-1), we actually already found that when finding our right sided limit. I know that that function value is just -1 + 1, which is equal to 0.

So with that function value of 0 here, I see that comparing my limits to my function value, these are definitely not the same thing. So I am dealing with a discontinuity here. For this piecewise function, it is discontinuous for x = -1. If you're having trouble visualizing what this may look like, because we're dealing with a piecewise function here and my first piece is just a constant whereas my second one is just a linear function, I am dealing with a jump discontinuity, meaning that I definitely could not trace this function without picking up my pen.

Now that we know how to determine where the discontinuities of a function are, let's continue practicing with this. Thanks for watching, and I'll see you in the next one.

Determine Continuity Algebraically Example 1

In this problem, we're asked to determine the values of \( x \), if any, for which this function is discontinuous. The function given here is \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \). Since this is a rational function, to determine where this function is discontinuous, we need to take the denominator and set it equal to 0. This will tell us anywhere that our rational function has holes or asymptotes, meaning that it would be discontinuous at those points.

So here, I take the denominator \( x^2 + 2x - 15 \), and I want to set that equal to 0. The easiest way to solve this quadratic equation and find those values of \( x \) which are 0 is to factor it. We want to look for two numbers that multiply to -15 and add to 2. Therefore, the factors are \( x - 3 \) and \( x + 5 \). These give us \( (x - 3)(x + 5) = 0 \). Since this is equal to 0, we can take each individual factor and set them equal to 0. Solving \( x - 3 = 0 \), by adding 3 to both sides, we find \( x = 3 \). This is one value of \( x \) for which this function is discontinuous. Solving \( x + 5 = 0 \), by subtracting 5 from both sides, we get \( x = -5 \). This is the other value for which the function is discontinuous.

Thus, this function is specifically discontinuous at \( x = 3 \) and \( x = -5 \). If you are curious about the type of discontinuities we have here, we can take a look back at our function. Since our numerator is \( x - 3 \), the same as one of the factors in the denominator, the \( x = 3 \) represents a removable discontinuity or a hole because it cancels out with the factor in the numerator. Meanwhile, the \( x + 5 \) factor that gives us our value of \( x = -5 \) represents an asymptote in our function. In summary, we have two types of discontinuities: one is a hole, and the other is an asymptote.

Let me know if you have any questions. I'll see you in the next one.

Determine the value(s) of $x$ (if any) for which the function is discontinuous.

$f\left(x\right)=\frac{x-4}{x^2-x-12}$

• A. $x=-4,x=3$
• B. $x=4,x=-3$
• C. $x=4$
• D. Function is continuous everywhere.

Determine Continuity Algebraically Example 2

Hey, everyone. In this problem, we're asked to determine the values of x, if any, for which this function is discontinuous. Since the function that we're given here is a piecewise function, that means that we want to look at the x values in between each piece. So let's go ahead and get started here. Now looking at our first two pieces, the x value in between these two pieces is x equals negative 2. That's where we transition from our first piece to our second piece. So that means that we want to explore x equals negative 2 and see if our limits and our function value are the same because if they are, then our function is continuous there. But if not, then we know we have a discontinuity. So let's go ahead and do that.

Looking at x equals negative 2, we're first going to take a look at our limit, so the limit of f of x as x approaches negative 2. Since this is a piecewise function, that means that we need to make sure that we look at our limit coming in from the left and from the right in order to accurately determine our limit. So here, we want to split this into our one-sided limits to see if these are the same. So looking at our left-sided limit first, limx→−2f(x) from the left. Looking up at my function here as we approach negative 2 from the left, I know that my function value is just going to be 5. So this is my left-sided limit. It's just 5. Now for our right-sided limit, limx→2f(x) from the right. Looking back up at my piecewise function, I know that coming in from the right side, my function is going to be x² + 2x + 1. So I need to go ahead and plug a negative 2 into this function in order to look at my right-sided limit. So plugging that in, I have (-2)² + 2(-2) + 1, which ends up giving me 4 - 4 + 1. Those fours cancel each other out, so my right-sided limit is 1.

Now here, my left-sided limit is 5, and my right-sided limit is 1. These are not the same value, which means that my regular limit, limx→−2f(x), definitely won't be equal to our function value f(-2). So we really don't even have to worry about finding that, but let's go ahead and find it just to cover all our bases. Now for f(-2), I know that my function is going to be equal to this quadratic function x² + 2x + 1. Now we actually already found what that is with our right-sided limit, so I know that f(-2) is 1. Now if my limit does not exist, that is definitely not the same thing as being 1. So here, since our limit and our function value are not the same, we know for sure that we have a discontinuity at x equals -2.

Now we do have another piece to our piecewise function, so we need to take a look at one other x value here. Now our other x value in between these two pieces is x equals 0. So let's go ahead and explore x equals 0 here. So, for x equals 0, we want to look at the same stuff. So let's start by taking a look at our limit. So we want to look at the limit of our function f of x as x approaches 0. Again, here, we want to look at our one-sided limits. So we're going to start with our left-sided limit, limx→0f(x) from the left. Now coming in from that left side, my function is going to be equal to that same quadratic function, x² + 2x + 1. So I can go ahead and plug 0 into that function in order to get my left-sided limit. So plugging 0 in, I have 0² + 2*0 + 1. Those first two terms end up being 0, so my left-sided limit is just 1.

Now let's look at our right-sided limit, limx→0f(x) from the right. Now coming in from the right side, my function is going to be equal to that last piece, x + 1. So I want to go ahead and plug 0 in to get that right-side limit to my function x + 1. Now plugging that in, I have 0+1, which we know is just 1. So my left-sided limit is 1. My right-sided limit is also 1. Because these are the same value, that tells me that my limit, my regular limit of f of x as x approaches 0 is also 1. So let's go ahead and check if this is equal to our function value. So we want to look at f(0). Now f(0), looking at my functions over here, is going to be equal to that same quadratic function, x² + 2x + 1. So I want to plug 0 into that function, which we actually have already done in that left-sided limit. So I already know that this is equal to 1.

Now because my limit here is equal to my function value, then my function is actually continuous at x equals 0. And I don't have a discontinuity here. So the only discontinuity in this piecewise function, having tested both of those x values in between my pieces, is going to be x equals negative 2. That is where this function is discontinuous. Now I know that that was a rather long and tedious process. Feel free to let me know if you have any questions here, and I'll see you in the next video.

Determine the value(s) of $x$ (if any) for which the function is discontinuous.

• A. $x=4,x=5$
• B. $x=5,x=1$
• C. $x=-1$
• D. Function is continuous everywhere