Continuity | Calculus - Mathematics PDF Download

Over the last few sections we’ve been using the term “nice enough” to define those functions that we could evaluate limits by just evaluating the function at the point in question. It’s now time to formally define what we mean by “nice enough”.
Definition
A function f(x) is said to be continuous at x=a if

A function is said to be continuous on the interval [a,b] if it is continuous at each point in the interval.
Note that this definition is also implicitly assuming that both  exist. If either of these do not exist the function will not be continuous at x=a.
This definition can be turned around into the following fact.

Fact I
If f(x) is continuous at x=a then,

This is exactly the same fact that we first put down back when we started looking at limits with the exception that we have replaced the phrase “nice enough” with continuous.
It’s nice to finally know what we mean by “nice enough”, however, the definition doesn’t really tell us just what it means for a function to be continuous. Let’s take a look at an example to help us understand just what it means for a function to be continuous.

Example 1 Given the graph of f(x), shown below, determine if f(x) is continuous at x=−2, x =0, and x=3.

Solution:
To answer the question for each point we’ll need to get both the limit at that point and the function value at that point. If they are equal the function is continuous at that point and if they aren’t equal the function isn’t continuous at that point.
First x = −2.

The function value and the limit aren’t the same and so the function is not continuous at this point. This kind of discontinuity in a graph is called a jump discontinuity. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite (i.e. the function doesn’t go to infinity).
Now x=0.

The function is continuous at this point since the function and limit have the same value.
Finally x=3.

The function is not continuous at this point. This kind of discontinuity is called a removable discontinuity. Removable discontinuities are those where there is a hole in the graph as there is in this case.
From this example we can get a quick “working” definition of continuity. A function is continuous on an interval if we can draw the graph from start to finish without ever once picking up our pencil. The graph in the last example has only two discontinuities since there are only two places where we would have to pick up our pencil in sketching it.
In other words, a function is continuous if its graph has no holes or breaks in it.
For many functions it’s easy to determine where it won’t be continuous. Functions won’t be continuous where we have things like division by zero or logarithms of zero. Let’s take a quick look at an example of determining where a function is not continuous.

Example 2 Determine where the function below is not continuous.

Solution:
Rational functions are continuous everywhere except where we have division by zero. So all that we need to is determine where the denominator is zero. That’s easy enough to determine by setting the denominator equal to zero and solving.
t² − 2t − 15 = (t−5)(t+3) = 0
So, the function will not be continuous at t=−3 and t=5.
A nice consequence of continuity is the following fact.

Fact 2
If f(x) is continuous at x=b and =b then,

To see a proof of this fact see the Proof of Various Limit Properties section in the Extras chapter. With this fact we can now do limits like the following example.

Example 3 Evaluate the following limit.

Solution:
Since we know that exponentials are continuous everywhere we can use the fact above.

Another very nice consequence of continuity is the Intermediate Value Theorem.
Intermediate Value Theorem
Suppose that f(x) is continuous on [a,b] and let M be any number between f(a) and f(b). Then there exists a number c such that,

1. a < c < b
2. f(c) = M

All the Intermediate Value Theorem is really saying is that a continuous function will take on all values between f(a) and f(b). Below is a graph of a continuous function that illustrates the Intermediate Value Theorem.

As we can see from this image if we pick any value, M, that is between the value of f(a) and the value of f(b) and draw a line straight out from this point the line will hit the graph in at least one point. In other words, somewhere between a and b the function will take on the value of M. Also, as the figure shows the function may take on the value at more than one place.
It’s also important to note that the Intermediate Value Theorem only says that the function will take on the value of M somewhere between a and b. It doesn’t say just what that value will be. It only says that it exists.
So, the Intermediate Value Theorem tells us that a function will take the value of M somewhere between a and b but it doesn’t tell us where it will take the value nor does it tell us how many times it will take the value. These are important ideas to remember about the Intermediate Value Theorem.
A nice use of the Intermediate Value Theorem is to prove the existence of roots of equations as the following example shows.

Example 4 Show that p(x)=2x³−5x²−10x+5 has a root somewhere in the interval [−1,2].
Solution:
What we’re really asking here is whether or not the function will take on the value
p(x)=0
somewhere between -1 and 2. In other words, we want to show that there is a number c such that −1<c<2 and p(c)=0. However if we define M=0 and acknowledge that a=−1 and b=2 we can see that these two condition on c are exactly the conclusions of the Intermediate Value Theorem.
So, this problem is set up to use the Intermediate Value Theorem and in fact, all we need to do is to show that the function is continuous and that M=0 is between p(−1) and p(2) (i.e. p(−1)<0<p(2) or p(2)<0<p(−1) and we’ll be done.
To do this all we need to do is compute,
p(−1)=8     p(2)=−19
So, we have,
−19=p(2)<0<p(−1)=8
Therefore M=0 is between p(−1) and p(2) and since p(x)
 is a polynomial it’s continuous everywhere and so in particular it’s continuous on the interval [−1,2]. So by the Intermediate Value Theorem there must be a number −1<c<2 so that,
p(c)=0or the sake of completeness here is a graph showing the root that we just proved existed. Note that we used a computer program to actually find the root and that the Intermediate Value Theorem did not tell us what this value was.

Let’s take a look at another example of the Intermediate Value Theorem.

Example 5 If possible, determine if f(x)=20 sin(x+3) cos  takes the following values in the interval [0,5].
(a) Does f(x)=10?
(b) Does f(x)=−10?
Solution:
Okay, so as with the previous example, we’re being asked to determine, if possible, if the function takes on either of the two values above in the interval [0,5]. First, let’s notice that this is a continuous function and so we know that we can use the Intermediate Value Theorem to do this problem.
Now, for each part we will let M be the given value for that part and then we’ll need to show that M lives between f(0) and f(5). If it does, then we can use the Intermediate Value Theorem to prove that the function will take the given value.
So, since we’ll need the two function evaluations for each part let’s give them here,
f(0)=2.8224     f(5)=19.7436
Now, let’s take a look at each part.

(a) Does f(x)=10?
Okay, in this case we’ll define M=10 and we can see that,
f(0)=2.8224<10<19.7436=f(5)
So, by the Intermediate Value Theorem there must be a number 0≤c≤5 such that
f(c)=10
(b) Does f(x)=−10?
In this part we’ll define M=−10. We now have a problem. In this part M does not live between f(0) and f(5). So, what does this mean for us? Does this mean that f(x)≠−10 in [0,5]
Unfortunately for us, this doesn’t mean anything. It is possible that f(x)≠−10 in [0,5], but is it also possible that f(x)=−10 in [0,5]. The Intermediate Value Theorem will only tell us that c’s will exist. The theorem will NOT tell us that c’s don’t exist.
In this case it is not possible to determine if f(x)=−10 in [0,5] using the Intermediate Value Theorem.
Okay, as the previous example has shown, the Intermediate Value Theorem will not always be able to tell us what we want to know. Sometimes we can use it to verify that a function will take some value in a given interval and in other cases we won’t be able to use it.
For completeness sake here is the graph of f(x) = 20 Sin(x + 3) cos   in the interval [0,5].

From this graph we can see that not only does f(x)=−10 in [0,5] it does so a total of 4 times! Also note that as we verified in the first part of the previous example f(x)=10 in [0,5] and in fact it does so a total of 3 times.
So, remember that the Intermediate Value Theorem will only verify that a function will take on a given value. It will never exclude a value from being taken by the function. Also, if we can use the Intermediate Value Theorem to verify that a function will take on a value it never tells us how many times the function will take on the value, it only tells us that it does take the value.

Practice problems: Continuity

1. The graph of f(x) is given below. Based on this graph determine where the function is discontinuous.

Solution:
Before starting the solution recall that in order for a function to be continuous at x=a both f(a) and   must exist and we must have,

Using this idea it should be fairly clear where the function is not continuous.
First notice that at x=−4 we have,

and therefore, we also know that  doesn’t exist. We can therefore conclude that f(x) is discontinuous at x=−4 because the limit does not exist.
Likewise, at x=2 we have,

and therefore, we also know that

doesn’t exist. So again, because the limit does not exist, we can see that f(x) is discontinuous at x=2.
Finally let’s take a look at x=4. Here we can see that,

and therefore, we also know that However, we can also see that f(4) doesn’t exist and so once again f(x) is discontinuous at x=4 because this time the function does not exist at x=4.
All other points on this graph will have both the function and limit exist and we’ll have  and so will be continuous.
In summary then the points of discontinuity for this graph are : x=−4, x=2 and x=4.

2. The graph of f(x) is given below. Based on this graph determine where the function is discontinuous.

Solution:
Before starting the solution recall that in order for a function to be continuous at x=a both f(a) and must exist and we must have,

Using this idea it should be fairly clear where the function is not continuous.
First notice that at x=−8 we have,

and therefore, we also know that
 We can also see that f(−8)=−3 and so we have,

Because the function and limit have different values we can conclude that f(x) is discontinuous at x = −8.
Next let’s take a look at x=−2 we have,

and therefore, we also know that  doesn’t exist. We can therefore conclude that f(x) is discontinuous at x=−2 because the limit does not exist.
Finally let’s take a look at x=6. Here we can see we have,

and therefore, we also know that

doesn’t exist. So, once again, because the limit does not exist, we can conclude that f(x) is discontinuous at x=6.
All other points on this graph will have both the function and limit exist and we’ll have and so will be continuous.
In summary then the points of discontinuity for this graph are : x=−8, x=−2 and x=6.

For problems 3 – 7 using only Properties 1 – 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the given function is continuous or discontinuous at the indicated points.

3. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) x=−1, (b) x=0, (c) x=3?

Solution:
(a) x = -1
Before starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Therefore, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
So, here we go.

So, we can see that  and so the function is continuous at x=−1.

(b) x = 0
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
Here is the work for this part.

So, we can see that  and so the function is continuous at x=0.

(c) x = 3
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a). Although there is also of course the problem here that f(3) doesn’t exist and so we couldn’t plug in the value even if we wanted to.
This also tells us what we need to know however. As noted in the notes for this section if either the function or the limit do not exist then the function is not continuous at the point. Therefore, we can see that the function is not continuous at x = 3.
For practice you might want to verify that,

and so  also doesn’t exist.

4. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) z=−2, (b) z=0, (c) z=5?

Solution:
(a) z = −2,
Before starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Of course, even if we had tried to plug in the point we would have run into problems as g(−2) doesn’t exist and this tell us all we need to know. As noted in the notes for this section if either the function or the limit do not exist then the function is not continuous at the point. Therefore, we can see that the function is not continuous at z=−2.
For practice you might want to verify that,

and so also doesn’t exist.

(b) z = 0,
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
Therefore, because we can’t just plug the point into the function, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
Here is the work for this part.

So, we can see that and so the function is continuous at z=0.

(c) z = 5?
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a). Although there is also of course the problem here that g(5) doesn’t exist and so we couldn’t plug in the value even if we wanted to.
This also tells us what we need to know however. As noted in the notes for this section if either the function of the limit do not exist then the function is not continuous at the point. Therefore, we can see that the function is not continuous at z = 5.
For practice you might want to verify that,

and so  also doesn’t exist.

5. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) x=4, (b) x=6?

Solution:
(a) x = 4,
Before starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Therefore, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
For this part we can notice that because there are values of x on both sides of x=4 in the range x<6 we won’t need to worry about one-sided limits here. Here is the work for this part.

So, we can see that
and so the function is continuous at x=4.

(b) x = 6?
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
For this part we have the added complication that the point we’re interested in is also the “cut-off” point of the piecewise function and so we’ll need to take a look at the two one sided limits to compute the overall limit and again because we are being asked to determine if the function is continuous at this point we’ll need to resort to basic limit properties to compute the one-sided limits and not just plug in the point (which assumes continuity again…).
Here is the work for this part.

So we can see that, does not exist.
Now, as discussed in the notes for this section, in order for a function to be continuous at a point both the function and the limit must exist. Therefore, this function is not continuous at x = 6 because

does not exist.

6. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) t = −2, (b) t = 10?

Solution:
(a) t=−2,
Before starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Therefore, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
Also notice that for this part we have the added complication that the point we’re interested in is also the “cut-off” point of the piecewise function and so we’ll need to take a look at the two one sided limits to compute the overall limit and again because we are being asked to determine if the function is continuous at this point we’ll need to resort to basic limit properties to compute the one-sided limits and not just plug in the point (which assumes continuity again…).
Here is the work for this part.

So we can see that
Next, a quick computation shows us that h(−2) = −2+6=4 and so we can see that
and so the function is continuous at t = −2.

(b) t = 10?
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
For this part we can notice that because there are values of t on both sides of t=10 in the range t ≥ −2 we won’t need to worry about one-sided limits here. Here is the work for this part.
Here is the work for this part.

So, we can see that and so the function is continuous at t = 10.

7. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) x = −6, (b) x = 1?

Solution:
(a) x = −6
Before starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Therefore, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
Also notice that for this part we have the added complication that the point we’re interested in is also the “cut-off” point of the piecewise function and so we’ll need to take a look at the two one sided limits to compute the overall limit and again because we are being asked to determine if the function is continuous at this point we’ll need to resort to basic limit properties to compute the one-sided limits and not just plug in the point (which assumes continuity again…).
Here is the work for this part.

So, we can see that, does not exist.
Now, as discussed in the notes for this section, in order for a function to be continuous at a point both the function and the limit must exist. Therefore, this function is not continuous at x=−6 because

(b) x = 1?
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
Again, note that we are dealing with another “cut-off” point here so we’ll need to use one-sided limits again as we did in the previous part.
Here is the work for this part.

So, we can see that,  and so  
Next, a quick computation shows us that g(1)=1 and so we can see that and so the function is continuous at x=1.

For problems 8 – 12 determine where the given function is discontinuous.

8. Determine where the following function is discontinuous.

Solution:
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since both are polynomials) the only points in which the rational expression will be discontinuous will be where we have division by zero.
Therefore, all we need to do is determine where the denominator is zero and that is fairly easy for this problem.

The function will therefore be discontinuous at the points :

9. Determine where the following function is discontinuous.

Solution:
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since both are polynomials) the only points in which the rational expression will be discontinuous will be where we have division by zero.
Therefore, all we need to do is determine where the denominator is zero and that is fairly easy for this problem.

The function will therefore be discontinuous at the points :

10. Determine where the following function is discontinuous.

Solution:
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since the numerator is just a constant and the denominator is a sum of continuous functions) the only points in which the rational expression will be discontinuous will be where we have division by zero.
Therefore, all we need to do is determine where the denominator is zero. If you don’t recall how to solve equations involving trig functions you should go back to the Review chapter and take a look at the Solving Trig Equations sections there.
Here is the solution work for determining where the denominator is zero. Using our calculator we get,

The second angle will be in the fourth quadrant and is 2π−1.0472=5.2360.
The denominator will therefore be zero at,

The function will therefore be discontinuous at the points,

Note as well that this was one of the few trig equations that could be solved exactly if you know your basic unit circle values. Here is the exact solution for the points of discontinuity.

11. Determine where the following function is discontinuous.

Solution:
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous (as we have here since the numerator is a polynomial and the denominator is a sum of two continuous functions) the only points in which the rational expression will be discontinuous will be where we have division by zero.
Therefore, all we need to do is determine where the denominator is zero and that is fairly easy for this problem.

The function will therefore be discontinuous at :

12. Determine where the following function is discontinuous.
g(x) = tan(2x)
Solution:
As noted in the hint for this problem when dealing with a rational expression in which both the numerator and denominator are continuous the only points in which the rational expression will not be continuous will be where we have division by zero.
Also, writing the function as,

we can see that we really do have a rational expression here. Therefore, all we need to do is determine where the denominator (i.e. cosine) is zero. If you don’t recall how to solve equations involving trig functions you should go back to the Review chapter and take a look at the Solving Trig Equations sections there.
Here is the solution work for determining where the denominator is zero. Using our basic unit circle knowledge we know where cosine will be zero so we have,

The denominator will therefore be zero, and the function will be discontinuous, at,

For problems 13 – 15 use the Intermediate Value Theorem to show that the given equation has at least one solution in the indicated interval. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval.

13. Use the Intermediate Value Theorem to show that 25−8x²−x³=0 has at least one root in the interval [−2,4]. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval,

Solution:
f(x) = 25 − 8x² − x³  &  M = 0
The problem is then asking us to show that there is a c in [−2,4] so that,
f(c) = 0 = M
but this is exactly the second conclusion of the Intermediate Value Theorem. So, let’s see that the “requirements” of the theorem are met.
First, the function is a polynomial and so is continuous everywhere and in particular is continuous on the interval [−2,4]. Note that this IS a requirement that MUST be met in order to use the IVT and it is the one requirement that is most often overlooked. If we don’t have a continuous function the IVT simply can’t be used.
Now all that we need to do is verify that M is between the function values as the endpoints of the interval. So,
f(−2) = 1    f(4) = −167
Therefore, we have,
f(4) = −167 < 0 < 1 = f(−2)
So, by the Intermediate Value Theorem there must be a number c such that,−2 < c < 4  &  f(c) = 0
and we have shown what we were asked to show.

14. Use the Intermediate Value Theorem to show that w²− 4ln(5w+2) = 0 has at least one root in the interval [0,4]. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval,

Solution:
f(w) = w² − 4ln(5w+2)  &  M = 0
The problem is then asking us to show that there is a c in [−2,4] so that,
f(c) = 0 = M
but this is exactly the second conclusion of the Intermediate Value Theorem. So, let’s see that the “requirements” of the theorem are met.
First, the function is a sum of a polynomial (which is continuous everywhere) and a natural logarithm (which is continuous on  - i.e. where the argument is positive) and so is continuous on the interval [0,4]. Note that this IS a requirement that MUST be met in order to use the IVT and it is the one requirement that is most often overlooked. If we don’t have a continuous function the IVT simply can’t be used.
Now all that we need to do is verify that M is between the function values as the endpoints of the interval. So,
f(0)=−2.7726    f(4)=3.6358
Therefore, we have,
f(0) = −2.7726 < 0 < 3.6358 = f(4)
So, by the Intermediate Value Theorem there must be a number c such that,
0 < c < 4  &  f(c) = 0
and we have shown what we were asked to show.

15. Use the Intermediate Value Theorem to show that 4t + 10e^t − e^2t = 0 has at least one root in the interval [1,3]. Note that you are NOT asked to find the solution only show that at least one must exist in the indicated interval,

Solution:
f(t) = 4t + 10e^t − e^2t  &  M = 0
The problem is then asking us to show that there is a c in [−2,4] so that,
f(c)=0=M
but this is exactly the second conclusion of the Intermediate Value Theorem. So, let’s see that the “requirements” of the theorem are met.
First, the function is a sum and difference of a polynomial and two exponentials (all of which are continuous everywhere) and so is continuous on the interval [1,3]. Note that this IS a requirement that MUST be met in order to use the IVT and it is the one requirement that is most often overlooked. If we don’t have a continuous function the IVT simply can’t be used.
Now all that we need to do is verify that M is between the function values as the endpoints of the interval. So,
f(1)=23.7938   f(3)=−190.5734
Therefore, we have,
f(3) = −190.5734 < 0 < 23.7938 = f(1)
So, by the Intermediate Value Theorem there must be a number c such that,
1 < c < 3  &  f(c) = 0
and we have shown what we were asked to show.