Lesson 1.8 · Unit 1 · CED topics 1.10 and 1.11
Types of discontinuities and continuity at a point
Informally, a function is continuous if you can draw its graph without lifting your pencil. That picture is a good start, but calculus needs a definition precise enough to prove things with. Limits provide it. In this lesson you will state exactly what continuity at a point means, test it step by step, and name the three types of break in a graph.
Learning objective.
- LIM-2.A. Justify conclusions about continuity at a point using the definition.
Essential knowledge.
- LIM-2.A.1. Types of discontinuities include removable discontinuities, jump discontinuities, and discontinuities due to vertical asymptotes.
- LIM-2.A.2. A function \(f\) is continuous at \(x=c\) provided that \(f(c)\) exists, \(\lim_{x\to c}f(x)\) exists, and \(\lim_{x\to c}f(x)=f(c)\).
Continuity at a point
A function \(f\) is continuous at \(x=c\) if all three of these conditions hold:
- \(f(c)\) exists (the function is defined at \(c\));
- \(\displaystyle\lim_{x\to c}f(x)\) exists (the left and right limits exist and agree); and
- \(\displaystyle\lim_{x\to c}f(x)=f(c)\) (the limit equals the function value).
In one line: \(f\) is continuous at \(c\) provided that \(f(c)\) exists, the limit as \(x\to c\) exists, and \(\displaystyle\lim_{x\to c}f(x)=f(c)\). If any condition fails, \(f\) is discontinuous at \(c\). When you are asked to justify continuity or discontinuity, check the conditions in order and say which one fails, if one does. "The graph has a hole" is a description; "\(\lim_{x\to2}f(x)=4\) but \(f(2)=3\)" is a justification.
Types of discontinuity
Discontinuities come in three kinds, according to which condition fails and how.
| Type | What happens | Which part fails |
|---|---|---|
| Removable | The limit exists, but \(f(c)\) is missing or has a different value (a hole) | \(f(c)\) undefined, or \(\lim\ne f(c)\) |
| Jump | The one-sided limits exist but are not equal | the two-sided limit does not exist |
| Infinite (vertical asymptote) | The function is unbounded near \(c\) | the limit does not exist (infinite) |
The word removable is a promise: since the limit exists, you could repair the graph by defining or redefining \(f(c)\) to equal the limit. Lesson 1.13 does exactly that. Jump and infinite discontinuities cannot be repaired by changing one value.
Let \(f(x)=\begin{cases}\dfrac{x^2-4}{x-2},&x\ne2\\ 3,&x=2.\end{cases}\) Is \(f\) continuous at \(x=2\)?
1. Does \(f(2)\) exist? Yes, \(f(2)=3\).
2. Does the limit exist? For \(x\ne2\), \(f(x)=\dfrac{(x-2)(x+2)}{x-2}=x+2\), so \(\displaystyle\lim_{x\to2}f(x)=4\). It exists.
3. Does the limit equal \(f(2)\)? \(4\ne3\). Condition 3 fails.
So \(f\) is not continuous at \(x=2\). Because the limit exists, the discontinuity is removable: redefining \(f(2)=4\) would make the function continuous.
Find and classify the discontinuities of \(h(x)=\dfrac{x+3}{x^2+x-6}\).
Factor the denominator: \(x^2+x-6=(x+3)(x-2)\), so \(h(x)=\dfrac{x+3}{(x+3)(x-2)}\). The function is undefined at \(x=-3\) and \(x=2\).
At \(x=-3\). For \(x\ne-3\), \(h(x)=\dfrac{1}{x-2}\), so \(\displaystyle\lim_{x\to-3}h(x)=\dfrac{1}{-5}=-\dfrac15\). The limit exists, so this is a removable discontinuity (a hole at \(\left(-3,-\frac15\right)\)).
At \(x=2\). The numerator is \(5\) and the denominator approaches \(0\), so \(|h(x)|\) grows without bound. This is an infinite discontinuity (a vertical asymptote at \(x=2\)).
Let \(p(x)=\begin{cases}x^2,&x\lt2\\ 8-x,&x\ge2.\end{cases}\) Show that \(p\) is discontinuous at \(x=2\) and classify the discontinuity.
\(p(2)=8-2=6\). Left limit: \(\displaystyle\lim_{x\to2^-}x^2=4\). Right limit: \(\displaystyle\lim_{x\to2^+}(8-x)=6\). The one-sided limits are not equal, so \(\displaystyle\lim_{x\to2}p(x)\) does not exist: condition 2 fails. This is a jump discontinuity, with a jump of \(6-4=2\).
Reading the conditions from a graph
On a graph, check by eye: Is there a filled-in point at \(x=c\)? (Condition 1.) Do the left and right sides of the curve arrive at the same height? (Condition 2.) Is the filled-in point at that height? (Condition 3.) If you can answer yes three times, the function is continuous there.
- Continuous at \(c\) means three things: \(f(c)\) exists, \(\lim_{x\to c}f(x)\) exists, and the two are equal. Say which condition fails when it is not.
- Removable: the limit exists but does not match \(f(c)\) or \(f(c)\) is undefined. Jump: one-sided limits differ. Infinite: vertical asymptote.
- A rational function has a removable discontinuity where a factor cancels and an infinite one where it does not.
- A limit existing at \(c\) does not make \(f\) continuous at \(c\): the function value must agree.
- Which condition for continuity fails for \(g(x)=\dfrac{1}{x-1}\) at \(x=1\), and what type of discontinuity is it?
Show answer
\(g(1)\) is undefined, and the limit does not exist (the function is unbounded near 1). It is an infinite discontinuity (vertical asymptote).
- Is \(f(x)=\dfrac{x^2-1}{x-1}\) continuous at \(x=1\)? Classify the discontinuity.
Show answer
\(f(1)\) is undefined, so \(f\) is discontinuous at 1. Because \(\dfrac{x^2-1}{x-1}=x+1\) for \(x\ne1\), the limit is 2. The discontinuity is removable.
- Let \(h(x)=\dfrac{|x|}{x}\) for \(x\ne0\). Classify the discontinuity at \(x=0\).
Show answer
The left limit is \(-1\) and the right limit is \(1\). They differ, so the two-sided limit does not exist: a jump discontinuity.
- Is \(f(x)=\begin{cases}x^2+1,&x\le2\\ 7-x,&x\gt2\end{cases}\) continuous at \(x=2\)? Justify with the three conditions.
Show answer
\(f(2)=5\). Left limit: \(2^2+1=5\). Right limit: \(7-2=5\). The limit exists and equals 5, which equals \(f(2)\). All three conditions hold, so \(f\) is continuous at 2.
- A graph has a closed circle at \((4,1)\), and the curve approaches height 3 from both sides of \(x=4\). Which condition fails? What type of discontinuity is it?
Show answer
\(f(4)=1\) exists and the limit is 3, but \(3\ne1\), so condition 3 fails. The discontinuity is removable.
- True or false: if \(\displaystyle\lim_{x\to c}f(x)\) exists, then \(f\) is continuous at \(c\). Explain.
Show answer
False. The limit may exist while \(f(c)\) is undefined or has a different value, as with a hole in the graph. Continuity needs the limit to exist and equal \(f(c)\).