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Continuity FAQs
Continuity 

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 nonremovable 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.
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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 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.
PatrickJMT  Continuity  
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.
PatrickJMT  Continuity  Limit Definition  
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
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.
Vertical Asymptote   This graph shows the equation \(\displaystyle{ f(x) = \frac{1}{x1} }\). 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. 

Hole   This plot shows the graph of the equation \(\displaystyle{ g(x)=\frac{x^21}{x1} }\). 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 \(x=1\). 

Jump   In this third plot, we are graphing
\(\displaystyle{
h(x) = \left\{
\begin{array}{rcl}
x+1 & & x \leq 1 \\
x^2 & & x > 1
\end{array}
\right.
}\)

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.
Calculus Expert  Discontinuities  
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.
\(n(x) \neq 0\)  \(n(x) = 0\)  Discontinuities?  

\(d(x) \neq 0\)  Zero  No  
\(d(x) = 0\)  Vertical Asymptote  Hole  Yes 
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Intermediate Value Theorem
Intermediate Value Theorem
The intermediate value theorem is used to establish that a function passes through a certain yvalue and relies heavily on continuity.
Intermediate Value Theorem 

For a continuous function, \(f(x)\) on an interval \([a,b]\), if \( t \) is between \(f(a)\) and \(f(b)\), 
Notice that the theorem just tells you that the value \(c\) exists but doesn't tell you what it is or how to find it.
To get the idea of this theorem clear in your head, here are some great videos for you to watch. They use graphs to help you understand what the theorem means.
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.
Calculus Expert  Intermediate Value Theorem  
Here is a great video that clearly explains the intermediate value theorem more from a mathematical point of view than in the previous video.
PatrickJMT  Intermediate Value Theorem  
Application of the Intermediate Value Theorem   Here is a great video showing a nonstandard 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.
Dr Chris Tisdell  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.
helpful? 4 

faq: What is a root of a function?
A root of a function is just a fancy word for an xintercept, i.e. it is where the graph crosses the xaxis. The term root is often used in electrical engineering. You can also call a root, a zero ( since \(y=0\) ).
synonyms 

root 
zero 
xintercept 
Note that you can have multiple roots of a function since a graph can cross with the xaxis 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.
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Practice Problems 

Instructions   Unless otherwise instructed, solve the following practice problems, giving your answers in exact, completely factored form.
Level A  Basic 
Practice A01  

Classify the discontinuities (hole, asymptote. . .) of \(\displaystyle{f(x)=\frac{x2}{x^24}}\) and determine which are removable.  
answer 
solution 
Practice A02  

Show that \(x^33x+1=0\) has a root in the interval \((0,1)\).  
solution 
Practice A03  

\(\displaystyle{
g(x) = \left\{
\begin{array}{lr}
x3 & x \leq 1 \\
x^2+1 & 1 < x \leq 2 \\
x^3+4 & x >2
\end{array}
\right.
}\)  
solution 
Practice A04  

Show that \(f(x)=x^4+x3\) has a root in the interval \((1,2)\).  
solution 
Practice A05  

\(\displaystyle{
f(x) = \left\{
\begin{array}{lr}
x^2+3x & x < 0 \\
\sqrt{x}+1 & x \geq 0
\end{array}
\right.
}\)
 
solution 
Practice A06  

\(\displaystyle{
f(x) = \left\{
\begin{array}{lr}
2x+1 & x \leq 1 \\
3x & 1 < x < 1 \\
2x1 & x \geq 1
\end{array}
\right.
}\)
 
solution 
Practice A07  

Show that \(x^2=\sqrt{x+1}\) has a root in the interval \((1,2)\).  
solution 
Practice A08  

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 
Level B  Intermediate 
Practice B01  

Show that \(e^x=2\cos(x)\) has at least one positive root.  
solution 
Practice B02  

\(\displaystyle{
f(x) = \left\{
\begin{array}{lr}
x^2+3 & x < 1 \\
ax+6 & x \geq 1
\end{array}
\right.
}\)
 
solution 
Practice B03  

\(\displaystyle{
f(x) = \left\{
\begin{array}{lr}
c^2  x^2 & x < 0 \\
2(xc)^2 & x \geq 0
\end{array}
\right.
}\)
 
solution 
Practice B04  

Find where \(\displaystyle{f(x)=\sqrt[3]{\frac{x+1}{x1}}}\) is continuous.  
solution 
Practice B05  

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