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. Before we get started, let's watch an introduction video to get an idea of where we are going.
video by Dr Chris Tisdell 

Definition of Continuity
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 \(x=c\). 
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. 
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 \(f(c)\). 
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 condition 2 above, where the limit does not exist, sometimes we need to look at onesided limits, i.e. limits from each side of the value we are talking about. You will find discussion, videos and practice problems on the onesided 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  \( (x+3)/(x^21) \) 
root functions  \( \sqrt{x+7} \) 
trig functions  \( \sin(x) \) 
logarithm functions  \( \ln(x1) \) 
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.
video by PatrickJMT 

2. This next video 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.
video by PatrickJMT 

Okay, now that you have a better understanding of continuity, take a look at discontinuities explained next. There is a great video in this section that will help you a lot. The discussion is going to cover several, seemingly diverse, topics. However, they are related in that the resulting equations are similar.
Discontinuities  Removable and Nonremovable
First let's discuss the 3 main types of discontinuities: vertical asymptotes, holes and jumps. Here are three graphs demonstrating each type.
\(\displaystyle{ f(x) = \frac{1}{x1} }\) 

1. Vertical Asymptote   For this graph, at \(x=1\) we have a vertical asymptote. This is a nonremovable discontinuity, which means that we can't redefine the 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 nonzero number in the numerator. The zero in the denominator removes the \(x=1\) from the domain and the nonzero number in the numerator tells us that we have an asymptote, consequently, a nonremovable discontinuity.
\(\displaystyle{ g(x)=\frac{x^21}{x1} }\) 

2. Hole   For this graph, 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.
3. Jump   In this third plot 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 precalculus piecewise function page.
\(\displaystyle{ h(x) = \left\{ \begin{array}{rcl} x+1 & & x \leq 1 \\ x^2 & & x > 1 \end{array} \right. }\) 

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

video by Krista King Math 

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 xaxis, 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 xvalues 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 yvalue 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 xvalues where the numerator AND denominator are both zero at the same xvalue. 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 xvalues 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 xvalues where \(n(x) = 0\) and \(d(x) \neq 0\)
 Holes occur at xvalues where \(n(x) = 0\) and \(d(x) = 0\)
 Vertical Asymptotes occur at xvalues where \(n(x) \neq 0\) and \(d(x) = 0\)
Here is the same information in table form.
\(f(x)=n(x)/d(x)\)  

\(n(x) \neq 0\) 
\(n(x) = 0\) 
Discontinuity?  
\(d(x) \neq 0\)  Zero  No  
\(d(x) = 0\)  Vertical Asymptote  Hole  Yes 
Practice
\(\displaystyle{ g(x) = \left\{ \begin{array}{lr} x3 & x \leq 1 \\ x^2+1 & 1 < x \leq 2 \\ x^3+4 & x >2 \end{array} \right. }\)
At what points is \(g(x)\) NOT continuous?
Problem Statement 

\(\displaystyle{ g(x) = \left\{ \begin{array}{lr} x3 & x \leq 1 \\ x^2+1 & 1 < x \leq 2 \\ x^3+4 & x >2 \end{array} \right. }\)
At what points is \(g(x)\) NOT continuous?
Solution 

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

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

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

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

\(\displaystyle{ f(x) = \left\{ \begin{array}{lr} 2x+1 & x \leq 1 \\ 3x & 1 < x < 1 \\ 2x1 & x \geq 1 \end{array} \right. }\)
At what points is \(f(x)\) NOT continuous?
Problem Statement 

\(\displaystyle{ f(x) = \left\{ \begin{array}{lr} 2x+1 & x \leq 1 \\ 3x & 1 < x < 1 \\ 2x1 & x \geq 1 \end{array} \right. }\)
At what points is \(f(x)\) NOT continuous?
Solution 

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

Find where \(\displaystyle{f(x)=\sqrt[3]{\frac{x+1}{x1}}}\) is continuous.
Problem Statement 

Find where \(\displaystyle{f(x)=\sqrt[3]{\frac{x+1}{x1}}}\) is continuous.
Solution 

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

Classify any discontinuities of the function \(\displaystyle{ f(x)=\frac{x2}{x^24} }\), then redefine the function at any removable discontinuities to make it continuous.
Problem Statement 

Classify any discontinuities of the function \(\displaystyle{ f(x)=\frac{x2}{x^24} }\), then redefine the function at any removable discontinuities to make it continuous.
Solution 

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

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

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

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

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

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

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

You CAN Ace Calculus
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
To bookmark this page and practice problems, log in to your account or set up a free account.
Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

Precalculus 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 