Check a piecewise join
Two formulas meet at one input
A piecewise function uses one formula on each side of an input. Here \(f(x) = x + 1\) when \(x < 2\), and \(f(x) = x^2 - 1\) when \(x \ge 2\). The input where the formula changes is the join, here \(x = 2\).
Press Close in from the left. An orange point moves along the left piece, and its height approaches 3.
Press Close in from the right. A second point moves along the right piece, and its height also approaches 3.
The filled dot at the join has height 3, because the rule \(x \ge 2\) gives \(f(2) = 2^2 - 1\). All three heights are equal, so the two pieces and the dot meet.
Three heights at a join
Without numbers, the left piece ends at a height \(\cy{L}\) and the right piece starts at a height \(\cy{R}\). The filled dot shows the value \(f(c)\).
Drag Lift the right piece. When \(R\) differs from \(L\), a red bar shows the gap between them. Drag Move the dot. When the dot leaves the right piece, a second red bar shows how far \(f(c)\) is from \(R\).
The function is continuous at the join only when \(L\), \(R\) and \(f(c)\) are all equal, which is when no red bar appears.
Notation
At the join \(c\), first find the left-hand limit. This is the height that the left piece approaches as \(x\) increases toward \(c\), so use only the formula for \(x < c\). If that formula is a polynomial, substitute \(c\) into it.
\[\lim_{x \to c^-} f(x) = \cy{L}\]Next, find the right-hand limit, the height that the right piece approaches as \(x\) decreases toward \(c\). Use only the formula for \(x > c\).
\[\lim_{x \to c^+} f(x) = \cy{R}\]Then find the value \(f(c)\). Read it from the line of the definition that includes \(x = c\).
\[f(c)\]The function is continuous at c when the three heights are equal.
\[\cy{L} = \cy{R} = f(c)\]If \(L = R\), the two-sided limit exists and equals this common height, so only the value \(f(c)\) can still break continuity.
When it fails
Press Lift the right piece. It starts higher than the left piece ends, so the one-sided limits differ and the two-sided limit does not exist.
No value at c can repair this, because the dot has one height and the pieces end at two different heights.
Example 1: pieces that meet
Let \(g(x) = 2x - 1\) for \(x < 3\), and let \(g(x) = x^2 - 4\) for \(x \ge 3\). Is \(g\) continuous at \(x = 3\)?
For the left-hand limit, use \(2x - 1\). Its height at \(x = 3\) is \(2(3) - 1 = 5\).
\[\lim_{x \to 3^-} (2x - 1) = \cy{5}\]For the right-hand limit, use \(x^2 - 4\). Its height at \(x = 3\) is also \(3^2 - 4 = 5\).
\[\lim_{x \to 3^+} (x^2 - 4) = \cy{5}\]The rule \(x \ge 3\) includes 3, so the value is the height of the right formula.
\[g(3) = 3^2 - 4 = 5\]All three heights are 5, so the two pieces meet at the filled dot and \(g\) is continuous at 3.
Example 2: pieces that meet, but not at the dot
Let \(h(x) = x^2 - 3x\) for \(x < 4\), let \(h(4) = 7\), and let \(h(x) = 2x - 4\) for \(x > 4\). Is \(h\) continuous at \(x = 4\)?
For the left-hand limit, use \(x^2 - 3x\). Its height at \(x = 4\) is \(16 - 12 = 4\).
\[\lim_{x \to 4^-} (x^2 - 3x) = \cy{4}\]For the right-hand limit, use \(2x - 4\). Its height at \(x = 4\) is \(8 - 4 = 4\).
\[\lim_{x \to 4^+} (2x - 4) = \cy{4}\]A tempting wrong move: Both sides approach 4, so \(h\) is continuous at 4.
\[L = R = 4 \Rightarrow \text{continuous}\]The filled dot is at height 7, not 4. The pieces meet each other at the ring, but a red bar of length 3 separates them from the dot.
The definition gives the value on its own line, \(h(4) = 7\). The limit is 4 but the value is 7, so \(h\) is not continuous at 4. The graph has a hole at \((4, 4)\) and a separate dot at \((4, 7)\).
\[4 \ne 7\]
Practice
Let \(p(x) = 3x - 2\) for \(x < 2\), and let \(p(x) = 10 - x^2\) for \(x \ge 2\). Is \(p\) continuous at \(x = 2\)? If not, how far apart are the heights?
Find the height that each piece reaches at \(x = 2\). The picture shows whether the pieces meet.
Solution
The left piece approaches the height \(3(2) - 2 = 4\).
\[\lim_{x \to 2^-} p(x) = \cy{4}\]The right piece starts at the height \(10 - 2^2 = 6\). The rule \(x \ge 2\) gives the value, so \(p(2) = 6\).
\[\lim_{x \to 2^+} p(x) = \cy{6} = p(2)\]The heights 4 and 6 differ, so \(p\) is not continuous at 2, and the graph jumps by \(6 - 4 = 2\).
Questions
Use the graph of f for the first two questions. A filled dot is a value. An open ring is a height that a piece approaches with no point there.
What is the left-hand limit at \(x = 1\)?
- 1
- 3
- 2
- The limit does not exist.
Answer
1
Why is \(f\) not continuous at \(x = 4\)?
- The left and right pieces approach different heights.
- Both pieces approach 2, but the value \(f(4)\) is 0.
- The function has no value at 4.
Answer
Both pieces approach 2, but the value \(f(4)\) is 0.
Let \(q(x) = 2x + 1\) for \(x < 3\), and let \(q(x) = x^2 - 1\) for \(x \ge 3\). By how much does the right-hand limit at \(x = 3\) exceed the left-hand limit?
- 1
- 0
- 7
- 8
Answer
1
Summary
You compare three heights at a join: the height that the left piece approaches, the height that the right piece approaches, and the value.
The function is continuous there when the three heights are equal, because then the two pieces and the dot meet.
This test applies at every input where a piecewise function changes formula.
This is one of four skills in topic 1.11 of Which & Why: AP Calculus.