Locate a discontinuity
Where does the pen lift?
The graph shows a function \(f\). Move the orange point along it. A pen that draws the graph must lift at some inputs.
At \(x = 2\) the graph has an open ring, so \(f\) has no value there and the pen lifts.
At \(x = 4\) the values on the left head to height 5 and the values on the right head to height 6, so the pen lifts.
At \(x = 6\) both sides head to height 4, but the value \(f(6)\) is the filled dot at height 1. The pen must lift to reach that dot.
At every other input the pen stays down. The graph breaks at 2, 4 and 6.
Three tests at one input
The graph now shows any function, and three tests apply at the input \(c\) of the orange point.
The first test asks whether \(f\) has a value at \(c\). The second asks whether the graph heads to the same height from the left and from the right. These heights are the left-hand and right-hand limits. The third test asks whether that common limit equals \(f(c)\).
A function is continuous at \(c\) when it passes all three tests. If it fails any test, \(f\) is discontinuous at \(c\), and the graph breaks there.
Move c along the graph. All three tests pass at most inputs, and at least one fails at each break.
Notation
First, \(f\) must have a value at \(c\). At \(x = 2\) the graph has an open ring, so this test fails.
\[\cy{f(c)}\ \text{exists}\]Second, the limit from the left and the limit from the right must exist and be equal. At \(x = 4\) they are 5 and 6, so this test fails.
\[\lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x)=\cy{L}\]Third, that common limit must equal the value. At \(x = 6\) the limit is 4 and the value is 1, so this test fails.
\[\cy{L}=f(c)\]At \(x = 3\) all three tests pass, so \(f\) is continuous there.
Test every input where the formula has no value or where one piece changes to another.
When it fails
Lower the value at \(x = 1\) with the button. The function still has a value there, and both sides still head to height 2, so the first two tests pass.
The limit 2 is not the value \(f(1)\), so the third test fails. One failed test is enough to make \(f\) discontinuous at 1.
Example 1: a formula with no value
Find the inputs where \(f(x)=\dfrac{x+2}{x^2-x-6}\) is discontinuous.
A fraction has no value where its denominator is 0, so factor the denominator.
\[x^2-x-6=(x-3)(x+2)\]The denominator is 0 at \(x = 3\) and at \(x = -2\), so \(f\) has no value at those two inputs. At every other input \(f\) has a value.
\[x=3,\qquad x=-2\]The first test fails at both inputs, so \(f\) is discontinuous at 3 and at \(-2\). For \(x\neq -2\), the factor \(x+2\) cancels, so the graph has a ring at \(-2\). At 3 nothing cancels, so the graph has a dashed vertical line.
\[\dfrac{x+2}{(x-3)(x+2)}=\dfrac{1}{x-3}\]
Example 2: where two pieces meet
Let \(g(x)=2x-1\) for \(x\leq 3\), and \(g(x)=x+1\) for \(x>3\). Find the inputs where \(g\) is discontinuous.
The formula changes at \(x = 3\), so test that input. The first piece covers \(x\leq 3\), so it gives the value, and the filled dot shows it.
\[g(3)=2(3)-1=5\]A tempting wrong move: The value \(g(3)=5\) exists, so \(g\) is continuous at 3.
\[g(3)=5\ \Rightarrow\ \text{continuous}\]A value passes only the first test. To the right of 3 the graph starts at height 4, so the pen must drop by 1, as the red segment shows.
From the left, \(2x-1\) heads to 5, and from the right, \(x+1\) heads to 4. The limits differ, so the second test fails.
\[\lim_{x\to 3^-} g(x)=\cy{5},\qquad \lim_{x\to 3^+} g(x)=\cy{4}\]The function \(g\) is discontinuous at 3. Each piece is a polynomial, and a polynomial has no breaks, so \(g\) has no other break.
Practice
Let \(p(x)=3x\) for \(x<2\), \(p(2)=5\), and \(p(x)=x^2+2\) for \(x>2\). Find the inputs where \(p\) is discontinuous.
Test the input where the formula changes. Compare \(p(2)\) with the limit from each side.
Solution
From the left, \(3x\) heads to 6. From the right, \(x^2+2\) also heads to 6, so the second test passes and the limit is 6.
\[3(2)=\cy{6},\qquad 2^2+2=\cy{6}\]The value \(p(2)=5\) is the dot below the ring. It is not 6, so the third test fails.
\[p(2)=5\neq 6\]The function \(p\) is discontinuous at 2. Each piece is a polynomial, so \(p\) has no other break.
Questions
The graph shows a function \(m\).
At which input does \(m\) have no value?
- \(-1\)
- 3
- 2
- \(m\) has a value at every input
Answer
\(-1\)
Which statement about \(x=3\) is correct?
- \(m(3)\) exists, but the left-hand and right-hand limits are different
- \(m(3)\) does not exist
- The two limits agree, but they differ from \(m(3)\)
- \(m\) is continuous at 3
Answer
\(m(3)\) exists, but the left-hand and right-hand limits are different
Let \(r(x)=\dfrac{x+1}{(x-2)(x+5)}\). At which inputs is \(r\) discontinuous?
- 2 and \(-5\)
- \(-1\), 2 and \(-5\)
- 2 only
- \(-1\) only
Answer
2 and \(-5\)
Summary
You list the inputs where the formula has no value or changes from one piece to another, and you test each one three ways.
This works because a continuous function has a value at the input, equal limits on both sides, and a limit equal to the value.
Use it to find every discontinuity of a graph or a formula before you name the kind of break.
This is one of five skills in topic 1.10 of Which & Why: AP Calculus.