Continuity - Online Tutor, Practice Problems & Exam Prep

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 the limit as x approaches c is the same as the function value.

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. First, let's take a look at our function on the left here. If we want to find the limit of this function as x approaches 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 is equal to 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 equals 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 approaches 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 is 2 here, I see that there's actually a hole on my graph when x is equal to 2. So the value of my function here is actually undefined. So here, my limit is definitely not the same as my function value. I know that this function is discontinuous for x equals 2.

By finding our limit and our function value, we're able to determine continuity using its definition. I also want to give you a visual tool to recognize continuity. If we take a look at our graph and take our pen or pencil and trace through our function, if we can trace through a specified point without stopping, then the function is continuous there. 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 is equal to 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 trace through my function now 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 2 because it is discontinuous there. You can think of this as a 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. With this function here, it was discontinuous at 2 specifically because there was a hole there. Discontinuities occur whenever a function has a hole but also when a function has either a jump or an asymptote.

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. With all of this in mind, let's go ahead and apply this knowledge to an example problem. 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. 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 equals negative 2. 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 is equal to negative 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 negative 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 negative 2. Well, let's move on to our next example here. Here we have c equals 4.

So we first want to find the limit here as x approaches 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. 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 is equal to 4, my function value is 1. 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. 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 equals 1. So looking at the limit of our function as x approaches 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. 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, specifically caused by our asymptote here. Now that we know what continuity is and how to find it, let's continue practicing. Thanks for watching, and I'll see you in the next one.

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

Intro to Continuity Example 1

In this problem, we're asked to state the intervals for which the function is continuous. So far, we've been identifying where our function is discontinuous. However, instead of that just being our answer, we want to state the intervals over which our function is continuous. We still need to figure out the same information.

We will use our knowledge of interval notation combined with our knowledge of function continuity to come up with this answer. Here is our graph of our function f(x), and it will be helpful to examine our graph going from left to right to determine our intervals of continuity. From the left side, we see that the function stretches out to negative infinity, indicating that my function starts from negative infinity as a continuous function. As we move towards the right until we reach a point of discontinuity at x=0, my first interval where my function is continuous is from negative infinity to 0.

Using our knowledge of interval notation, there is a solid dot at 0, which means that this first interval from negative infinity to 0 should be enclosed on that side with a square bracket. Now, let's continue moving to the right to see our next interval. Starting from the zero point, here I see an open circle, indicating that we should use a curved parenthesis and not a square bracket. The interval of continuity stretches from 0 until we reach the next point of discontinuity at x=3 due to an asymptote. Since it's an asymptote, my function will never actually be equal to 3; it will just get really, really close.

Thus, I need to use a parenthesis at 3. The second interval of continuity runs from 0 to 3. Continuing from the other side of this asymptote at 3, that is my next interval, starting at 3 again enclosed with a parenthesis because my x value is never actually equal to 3, it is just really close to it. Continuing to the right, I reach another discontinuity quickly at x=4.

This is the end of my next interval. Here, I see an open circle, so I need to use a parenthesis. Continuing from the point of 4, we pick back up again using a parenthesis as it is an open circle. Our function continues to extend to infinity from this side, so our final interval goes from 4 to positive infinity, using a parenthesis instead of a square bracket.

These are all of the intervals for which my function is continuous, and all the points in between represent where I have a discontinuity. We're just stating our answer differently. Let me know if you have any questions here, and thanks for watching.

Intro to Continuity Example 2

In this problem, we're asked to state the intervals for which the function f(x) is equal to the square root of \(2x - 1\) is continuous. Now we're given a graph here so that we can clearly see visually where this function is continuous. Remember that when stating intervals, we're going to be using our interval notation of square brackets or parentheses based on what's happening with our function. Here, we can see that our function starts right here at \(x = 0.5\) and continues on to infinity with no breaks, no asymptotes; it just continues on. So our function is actually continuous starting from that 0.5 using a square bracket because this is a solid point all the way to infinity.

And of course, for infinity, we use just a parenthesis. So this is the only interval of our function, it's actually its domain because it's continuous over its entire domain. Let me know if you have any questions here. Thanks for watching.

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. 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. That means that we need to be able to determine where these discontinuities are when working with both 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. Let's first 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 are given here is \( f(x) = \frac{x^2 - x - 6}{x + 2} \). 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. So 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 \( x + 2 \). So 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 \( x = -2 \). And this is where my function is discontinuous.

Now one of two 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. Now, it's not likely that you'll be asked often to figure out what type of discontinuity you're dealing with. But if you're curious here, for this function, you would actually end up with a hole or removable discontinuity. So on a graph, you would have to stop, pick up your pen, and continue on the other side of that hole.

Now that's for a rational function, but let's go ahead and move on to our piecewise function. Now here we want to do the same thing. We want to determine the values of x, if any, for which our function is discontinuous. Now here our function is equal to 4 for \( x < -1 \), and it's equal to \( x + 1 \) for \( x \geq -1 \). 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. Now here, since I only have two pieces, I only have to worry about one point here, and that's \( x = -1 \). And 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 \( f(-1) \). Because 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. So let's go ahead and find these values starting first with our limit. Now because this is a piecewise function, I need to be careful here and look at my one-sided limits. So I'm going to start with my left-sided limit, the limit of \( f(x) \), as \( x \) approaches \( -1 \) from that left side. Now looking at my piecewise function, I know that coming in from the left side, my function is going to be equal to 4 for these values of \( x < -1 \).

So I know that that left-sided limit is simply 4. Now my limit from the other side, my 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 that this limit is going to be equal to \( -1 + 1 \), which is simply 0, so that right-sided limit is 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, that tells me that my regular limit, the limit of \( f(x) \) as \( x \) approaches \( -1 \), simply does not exist. Now because this limit does not exist, I sort of already 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 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. So for this piecewise function, it is discontinuous for \( x = -1 \). Now 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 here, 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 the interval(s) for which the function is continuous.

$f(x)=x^2+4x-12$

• A. $(-∞,-12),(-12,∞)$
• B. $(-∞,-6),(-6,2),(2,∞)$
• C. $(-∞,∞)$

Determine Continuity Algebraically Example 3

In this problem, we're asked to determine the values of x, if any, for which this function is discontinuous. Now the function that we're given here is \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \). Since this is a rational function, in order to determine where this function is discontinuous, we want to take the denominator and set it equal to zero. 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 my denominator \( x^2 + 2x - 15 \), and I want to set that equal to zero.

Now the easiest way to solve this quadratic equation and find those values of x which are zero is to go ahead and factor here. Now we want to look for two numbers that multiply to negative 15 and add to positive 2. So that's going to be \( x - 3 \) and \( x + 5 \). That's going to give us our factors here. And that is still equal to zero here.

Now because this is equal to zero, that means that I can take each individual factor and set that equal to zero. So if I take \( x - 3 = 0 \) and \( x + 5 = 0 \) and solve for zero for each of these terms, If I add 3 on this side, I end up with \( x = 3 \). So that is one of my values of x for which this function is discontinuous. And then solving over here, if I subtract 5 on both sides ending up with \( x = -5 \), this is the other value for which my function is discontinuous. So, this function is specifically discontinuous at \( x = 3 \) and \( x = -5 \).

Now if you're curious about what type of discontinuities we're dealing with here, we can take a look back at our function. And since we see that our numerator is this \( x - 3 \), the same as one of our factors in the bottom, \( x = 3 \) represents a removable discontinuity or a hole because it cancels out with that top factor. Whereas our \( x + 5 \) factor that gives us our value of \( x = -5 \), this represents an asymptote in our function. So here we have 2 discontinuities: one is a hole, one is an asymptote.

Let me know if you have any questions here. 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 the interval(s) for which the function is continuous.

$f\left(x\right)=\frac{\sin x}{(2+\cos x)}$

• A. $(-1,1)$
• B. $(-∞,-2),(-2,∞)$
• C. $(-∞,∞)$
• D. The function is not continuous anywhere.

Determine Continuity Algebraically Example 4

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 we want to look at the x values between each piece. So let's go ahead and get started here. Now looking at our first two pieces, the x value 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 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, we need to make sure to look at our limit coming in from the left and from the right 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, lim fx as x→ -2 from the left. Looking 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, lim fx as x→ -2 from the right. Looking back at my piecewise function, I know that coming in from the right side, my function is going to be x2+2∙x+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 -22+2∙-2+1.

Now this ends up giving me 4 minus 4 plus 1. Those fours cancel each other out. So my right-sided limit is 1. Now here, my left-sided limit is 5. My right-sided limit is 1.

These are not the same value, which means that my regular limit, my limit of f of x as x approaches negative 2, definitely won't be equal to our function value f of negative 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 of negative 2, I know that my function is going to be equal to this quadratic function x2+2∙x+1. Now we actually already found what that is with our right-sided limit, so I know that f of negative 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 negative 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 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. We want to look at the limit and the function value. 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. Now, again, here, we want to look at our one-sided limit.

We're going to start with our left-sided limit, the limit of f of x as x approaches 0 from the left. Now coming in from that left side, my function is going to be equal to that same quadratic function, x2+2∙x+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, which is just 1.

Now let's look at our right-sided limit. The limit of f of x as x approaches 0 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-sided limit to my function x + 1.

Now plugging that in, I have 0+1, which 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 of 0. Now f of 0, looking at my functions over here, is going to be equal to that same quadratic function, x2+2∙x+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 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

Determine the interval(s) for which the function is continuous.

• A. $[-3,0),[0,\infty )$
• B. $(-∞,0),[0,∞)$
• C. $(-∞,∞)$
• D. The function is not continuous anywhere.