## Continuity

### Video List

ivt   apply the intermediate value theorem

B01  using the ivt

exp   explaining continuity

lim   limit definition of continuity

ivt   A01  A02  explain the intermediate value theorem and examples

A03  continuity with a piecewise function

A05  A06  discontinuities in piecewise functions

B02  making a piecewise function continuous

ivt   overview of the intermediate value theorem

A04  using ivt to prove a root

B05  prove a root

dsc   explain discontinuities

A07  determine removable discontinuities

B03  making a function continuous

B04  where a function is continuous

### Search 17calculus

Continuity is something best learned from graphs to get a feel for it. Then you can go to equations to cement the concept in your head.

There are 3 parts to continuity. At this moment we are talking about continuity at a point.
For a function f(x) to be continuous at a point x = c all three of these conditions must hold.

 1. $$f(c)$$ is defined. 2. $$\displaystyle{\lim_{x \rightarrow c}{f(x)}}$$ exists. 3. $$\displaystyle{\lim_{x \rightarrow c}{f(x)} = f(c)}.$$

If any one of these conditions is broken, then the function is not continuous at $$x=c$$.
Let's look at graphs where each of the above conditions does not hold.

Condition 1: $$f(c)$$ is not defined.
This graph shows an example of where the function is not defined at $$x=c$$. So this function is not continuous at $$x=c$$.

This is an example of a removable discontinuity. We can just redefine the function by redefining one point at $$f(c)$$ to make it continuous.

Condition 2: $$\displaystyle{\lim_{x \rightarrow c}{f(x)}}$$ does not exist.
Notice that the limit from the left is different than the limit from the right ( at $$x=c$$ ). This means the limit does not exist.

This is an example of a non-removable discontinuity at $$x=c$$. There is not way to 'plug the hole' or redefine the function at only one point so that the result is continuous.

Condition 3: $$\displaystyle{\lim_{x \rightarrow c}{f(x)} \neq f(c)}.$$
This graph shows an example of where the first two cases hold but the third doesn't, i.e. $$f(c)$$ is defined, the limit exists but the limit does not equal $$f(c)$$.

This is also an example of a removable discontinuity. Notice you can just move the $$f(c)$$ to fill the hole to make the function continuous.

Notes
1. Although not explicitly stated above, continuity holds in both directions, i.e. if a function is continuous then all three conditions hold and if all three conditions hold, then the function is continuous. So we can say, $$f(x)$$ is continuous at $$x=c$$ if and only if all three conditions listed above hold.
2. For case 2 above, where the limit must exist, sometimes we need to look at one-sided limits, i.e. limits from each side of the value we are talking about. You will find discussion, videos and practice problems on the one-sided limits page for this case.

Example Functions - - There are some functions that are guaranteed to be continuous on their domains. This is important . . . these functions are not necessarily continuous everywhere but they are continuous on their domains. We can use this information to build continuity information about other functions.

function type

example

polynomials

$$x^3+3x^2+1$$

rational functions

$$(x+3)/(x^2-1)$$

root functions

$$\sqrt{x+7}$$

trig functions

$$sin(x)$$

logarithm functions

$$ln(x-1)$$

Okay, now that you have an intuitive idea of continuity, let's watch some videos to help you understand and use continuity. It is important to watch both of them to get a complete picture of continuity.

1. This video explains continuity from a more mathematical viewpoint.

2. This is a great video to watch to get a much better understanding of continuity using the limit definition. This is one of the best videos we've seen on youtube that explains a complicated math topic in a way that is understandable.

Okay, now that you have a better understanding of continuity, take a look at discontinuities explained in the next panel. There is a great video in this section that will help you a lot.

### Discontinuities, Removable vs. Nonremovable; Zeroes, Holes and Asymptotes

This discussion is going to cover several, seemingly diverse, topics. However, they are related in that the resulting equations are similar.

Discontinuities --- Removable vs. Nonremovable
First let's discuss the 3 main types of discontinuities: jumps, holes and asymptotes. Here are three graphs demonstrating each type.

This graph shows the equation $$\displaystyle{ f(x) = \frac{1}{x-1} }$$. At $$x=1$$ we have a vertical asymptote. This is a nonremovable discontinuity, i.e. we can't redefine that function at x=1 with one value that will make the function continuous there.

Notice in the equation that at $$x=1$$, we have a zero in the denominator and a number (not zero) in the numerator.

This plot shows the graph of the equation $$\displaystyle{ g(x)=\frac{x^2-1}{x-1} }$$. At $$x=1$$ we have a hole. This is a removable discontinuity since if we add the single point (1,2) to the function, the result is a continuous function at $$x=1$$.

Notice in the equation of g(x) that at $$x=1$$, we have zero in the numerator and the denominator.

In this third plot, we are graphing $$\displaystyle{ h(x) = \left\{ \begin{array}{rcl} x+1 & & x \leq 1 \\ x^2 & & x > 1 \end{array} \right. }$$
At $$x=1$$ we have a jump. This is a nonremovable discontinuity since we can't redefine the function a single point to make it continuous at $$x=1$$. Note: This graph is a piecewise function. For a review, go to the piecewise function page.

Here is a great video explaining discontinuities in more detail with lots of examples. Her example of what she calls a crazy graph is especially good. As an instructor, I have often put questions like this on exams.

Zeroes, Holes and Asymptotes
Now let's look at the equations for each of these. I have included this with the discussion of discontinuities since two of these are discontinuities and the third, zeroes, are related to the other two but are not discontinuities. For this discussion, we are going to look how the equations are similar.

First let's look at zeroes. Zeroes of a function are sometimes called poles (especially in electrical engineering). Basically, they are points where the graph of a function crosses the x-axis, i.e. where $$y=0$$. They are not discontinuities but are important points in mathematics and engineering. If you have a function that is a fraction such as $$\displaystyle{ f(x)=\frac{n(x)}{d(x)} }$$, zeroes occur at x-values where the numerator function is zero but the denominator function is NOT zero. For example, look at the second graph above. A zero occurs at $$x = -1$$. You can also say that there is a zero at the point $$(-1,0)$$. By definition, the y-value is zero, so we usually do not write the point $$(-1,0)$$. We usually just say $$x=-1$$ or at $$-1$$.

Okay, let's look at holes. If we have the a function in fraction form that looks like $$\displaystyle{ f(x)=\frac{n(x)}{d(x)} }$$, holes occur at x-values where the numerator AND denominator are both zero at the same x-value. A hole is a discontinuity. Looking at the second graph above, we have a hole at $$x=1$$ because the numerator and denominator of g(x) are both zero at $$x=1$$.

Finally, vertical asymptotes occur at x-values where the denominator is zero but the numerator is NOT zero. Asymptotes are discontinuities. The first graph above shows this case. Notice that at $$x=1$$, the numerator of f(x) is 1, but the denominator is zero.

Let's sum this up. For a fractional function in the form $$\displaystyle{ f(x)=\frac{n(x)}{d(x)} }$$:
- Zeroes occur at x-values where $$n(x) = 0$$ and $$d(x) \neq 0$$
- Holes occur at x-values where $$n(x) = 0$$ and $$d(x) = 0$$
- Vertical Asymptotes occur at x-values where $$n(x) \neq 0$$ and $$d(x) = 0$$
Here is the same information in table form.

$$n(x) \neq 0$$ $$n(x) = 0$$ Discontinuities? Zero No Vertical Asymptote Hole Yes

Additional Information - - The pre-calculus page contains videos discussing graphing of rational functions.

### Intermediate Value Theorem

The intermediate value theorem is used to establish that a function passes through a certain y-value and relies heavily on continuity. It doesn't tell you what that y-value is, it just establishes it's existence.

Here is a video that shows, graphically, how the intermediate value theorem works. She uses color in her graph to make it easy to follow.

Here is a great video that clearly explains the intermediate value theorem more from a mathematical point of view than in the previous video.

Application of the Intermediate Value Theorem - - Here is a great video showing a non-standard application of the IVT. To work this problem, he uses the definition of the limit. Don't skip this video. It will help you understand limits, continuity and the IVT.

### faq: If a function has a nonremovable discontinuity at x = c, does x = c have to be a vertical asymptote?

#### If a function has a nonremovable discontinuity at x = c, does x = c have to be a vertical asymptote?

No, here is an example of a nonremovable discontinuity at $$x=c$$ that is not a vertical asymptote. Click here for more information about nonremovable discontinuities.

### faq: What is a root of a function?

A root of a function is just a fancy word for an x-intercept, i.e. it is where the graph crosses the x-axis. The term root is often used in electrical engineering. You can also call a root, a zero ( since $$y=0$$ ).

synonyms

root

zero

x-intercept

Note that you can have multiple roots of a function since a graph can cross with the x-axis multiple times without failing the vertical line test. [ which is why we talk about A root and not THE root ]

We discuss more about roots and how to determine if a function has a root from the equation in the panel on discontinuity above.

use these filters to show only the practice problems you want to see

continuity practice

intermediate value theorem practice

Practice Problems

 Level A - Basic

Practice A01

Show that $$x^3-3x+1=0$$ has a root in the interval $$(0,1)$$.

Practice A02

Show that $$x^2 = \sqrt{x+1}$$ has a root in the interval $$(1,2)$$.

Practice A03

$$\displaystyle{ g(x) = \left\{ \begin{array}{lr} x-3 & x \leq -1 \\ x^2+1 & -1 < x \leq 2 \\ x^3+4 & x >2 \end{array} \right. }$$
At what points is $$g(x)$$ NOT continuous?

Practice A04

Show that $$f(x)=x^4+x-3$$ has a root in the interval $$(1,2)$$.

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##### Practice Problem [ A03 ] Solution

$$g(x)$$ is not continuous at $$x=-1$$ and $$x=2$$

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##### Practice Problem [ A04 ] Solution

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Practice A05

$$\displaystyle{ f(x) = \left\{ \begin{array}{lr} x^2+3x & x < 0 \\ \sqrt{x}+1 & x \geq 0 \end{array} \right. }$$
At what points is $$f(x)$$ NOT continuous?

Practice A06

$$\displaystyle{ f(x) = \left\{ \begin{array}{lr} 2x+1 & x \geq -1 \\ 3x & -1 < x < 1 \\ 2x-1 & x \geq 1 \end{array} \right. }$$
At what points is $$f(x)$$ NOT continuous?

##### Practice Problem [ A05 ] Solution

$$f(x)$$ is not continuous at $$x = 0$$

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##### Practice Problem [ A06 ] Solution

$$f(x)$$ is not continuous at $$x=-1$$ and $$x=1$$

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Practice A07

$$\displaystyle{ f(x) = \frac{x-2}{x^2-4} }$$ Classify the discontinuities (hole, asymptote. . .) of $$f(x)$$ and determine which are removable.

##### Practice Problem [ A07 ] Solution

$$f(x)$$ has discontinuites
$$x=-2$$ which is an asymptote and nonremovable, and
$$x=2$$ which is a hole and is removable

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 Level B - Intermediate

Practice B01

Show that $$e^x = 2cos(x)$$ has at least one positive root.

Practice B02

$$\displaystyle{ f(x) = \left\{ \begin{array}{lr} x^2+3 & x < 1 \\ ax+6 & x \geq 1 \end{array} \right. }$$
What value of a is the piecewise function $$f(x)$$ continuous on the entire real line?

Practice B03

$$\displaystyle{ f(x) = \left\{ \begin{array}{lr} c^2 - x^2 & x < 0 \\ 2(x-c)^2 & x \geq 0 \end{array} \right. }$$
What value of c is the piecewise function $$f(x)$$ continuous on the entire real line?

##### Practice Problem [ B01 ] Solution

He chose $$x=0$$ and $$x=\pi/2$$ to show that the function has a positive value and negative value. You can choose other values as long as one results in a positive value and the other is negative value.

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##### Practice Problem [ B02 ] Solution

$$a = -2$$

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##### Practice Problem [ B03 ] Solution

$$c = 0$$

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Practice B04

$$\displaystyle{ f(x) = \sqrt[3]{\frac{x+1}{x-1}} }$$ Find where f(x) is continuous.

Practice B05

Prove that $$cos(x)=x^3$$ has at least one real root.

##### Practice Problem [ B04 ] Solution

$$f(x))$$ is continuous for $$x \neq 1$$

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