Continuity - Online Tutor, Practice Problems & Exam Prep

Intro to Continuity

Hey, everyone. Here, we're going to talk about continuity. We've already been dealing with continuity throughout our last several lesson videos. So here, we're really just putting a name to it. A function is continuous at some point c if lim→xc 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, we're going to dive a bit deeper into continuity and also talk about what it means for a function to be discontinuous. Let's first take a look at our function on the left here. To find the limit of this function as xc→2, as x gets really, really close to 2 from either side, my function is approaching a y-value of 4.

Now, looking at the actual value of my function when x is equal to 2, it's actually the exact same as my limit; it's also 4. So here, my limit is the same as my function value. This function is continuous at x2. Now, examining our other function, to find the limit as x2→2, it's approaching a y-value of 1.

However, the actual value of the function when x is 2 shows a hole on my graph, thus it's undefined. The limit is definitely not the same as the function value, indicating this function is discontinuous for x2.

By finding our limit and our function value, we're able to determine continuity using its definition. Using a visual tool, if we trace through our graph without stopping at a specified point, then the function is continuous there. For our parabola, as we trace through the function at x2, it's continuous; but another function shows discontinuity as we must stop and continue on the other side of two because of a hole.

Discontinuities occur for rational functions and piecewise functions. For our example problem, we find if functions are continuous at each given value of c by comparing the limit and function value. If these values are the same, the function is continuous; if not, it's discontinuous. For cnegative 2, we find both values are 1, indicating continuity. For c equals 4, the limit does not exist and the function value is 1, showing discontinuity.

Finally, for c equals 1, our function going down to negative infinity signifies an unbounded behavior where the limit does not exist, and the function value is undefined due to an asymptote, confirming discontinuity. With this understanding of continuity, 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 just need to state our answer differently using interval notation. Here, we are going to use our knowledge of interval notation combined with our knowledge of function continuity to come up with this answer. Now, here is our graph of our function f(x), and in order to determine our intervals of continuity, it's going to be helpful to look at our graph going from left to right. From this left side, we see that we have that function stretching out to negative infinity, which means that my function starts from negative infinity as a continuous function. Then, as we move towards the right until we reach a point of discontinuity, which is here at \( x = 0 \), my first interval that my function is continuous over is from negative infinity until we reach that 0.

Using our knowledge of interval notation, I do have a solid dot here at 0, which means that this first interval from negative infinity to 0 should be enclosed on that side with a square bracket. So that's my first interval of continuity. Let's continue moving further to the right to see our next interval. Starting from that zero point, that's my next interval. Here I see I have this open circle, meaning that I just need to use a curved parenthesis and not a square bracket. Let's see how far this interval of continuity goes.

From 0 until I reach my next point of discontinuity, which is here at \( x = 3 \) because I have an asymptote. So, from 0 to 3 is my next interval. Should I use a parenthesis or a square bracket here? Well, since this is an asymptote, that means that my function is never actually going to be equal to 3. It's just going to get really, really close to it.

So, that means that I need to use another parenthesis here. So that's my second interval of continuity. Let's continue on further. Now seeing where my graph picks back up on the other side of this asymptote at 3, that is my next interval starting at 3 again enclosed here with a parenthesis because my x value is not ever equal to 3, it is just really close to it. Now continuing on here, going more to the right, I reach another discontinuity rather quickly here at \( x = 4 \).

So this is the end of my next interval here. Should I use a square bracket or a parenthesis? Here, I see that I have this open circle, so that means I need to use a parenthesis. Now let's continue on and finish this up. Continuing from that point of 4, we're picking back up again using a parenthesis there because it is an open circle.

Then continuing on, our function does continue on to infinity from this side so our final interval goes from 4 to positive infinity and since we have an infinity there, we are going to use a parenthesis and not a square bracket. These are all of my intervals for which my function is continuous and you can see that all those points in between represent where I have a discontinuity. So again, we want to figure out the same information. We're just stating our answer differently. Feel free to ask 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 \( \sqrt{2x - 1} \) 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 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. And 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. 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. 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. Now the first function that we are given here is \( f(x) = \frac{x^2 - x - 6}{x+2} \). Now 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 actually would end up with a hole or a movable 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 less than negative one, and it's equal to \( x + 1 \) for x greater than or equal to negative one.

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. Now, 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 negative 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 negative one 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 less than negative one. 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 negative one 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 negative 1. So that means that this limit is going to be equal to negative 1 plus 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 negative one, 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 actually already found that when finding our right sided limit.

I know that that function value is just negative one plus 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.

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

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

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 \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \). Now because this is a rational function, we need to identify points where our rational function has holes or asymptotes, meaning that it would be discontinuous at those points. To do this, I take the denominator \( x^2 + 2x - 15 \), and set that equal to zero to find points of discontinuity.

Now, we want to look for two numbers that multiply to -15 and add to +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 0 here. Now because this is equal to 0, I can take each individual factor and set that equal to 0.

So if I take \( x - 3 = 0 \) and \( x + 5 = 0 \) and solve for \( x \) for each of these terms, adding 3 to one side, I end up with \( x = 3 \). This is one of my values of \( x \) for which this function is discontinuous. And then solving over here, subtracting 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 types of discontinuities we're dealing with here, we can take a look back at our function. Since we see that our numerator is the same \( x - 3 \) 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 types of discontinuities, one is a hole, and the other 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. $\text{}x=4\text{}$
• D. Function is continuous everywhere.

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

$f(x)=\frac{sin⁡x}{(2+cos⁡x)}$

• 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. Now, since the function that we're given here is a piecewise function, that means that 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. 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.

Now, because 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, the limit of f of x as x approaches negative 2 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, the limit of f of x as x approaches negative 2 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.

Now this 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. My right sided limit is 1.

These are not the same value, which means that my regular limit, the limit of f of x as x approaches a negative 2, actually does not exist. Now because we already know that our limit does not exist here, this 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 + 2x + 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 limits. So 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 + 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, 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-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 of 0. Now f of 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 between my pieces, is going to be x equals negative 2. That is where this function is discontinuous. Now I know 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.