Decide whether both sides agree
Two breaks, two answers
The graph has two breaks, at \(x = 1\) and at \(x = 3\). Press Compare at x = 1. Two orange points move toward 1, one from each side.
The point on the left follows the rising line and heads toward height 2. The point on the right follows the falling line and heads toward height 3. The two sides give different heights.
Press Compare at x = 3. The point on the left heads toward height 1, and the point on the right also heads toward height 1. At \(x = 3\) the two sides agree.
The filled dot at \((3, 3)\) is the value of \(f\) at 3, which this comparison does not use.
Any break
Two orange points close in on \(c\). The left part heads toward a height \(\cy{L}\), and the right part heads toward a height \(\cy{R}\). The orange bar between the two rings shows how far apart they are.
Move the slider R minus L. When \(R\) differs from \(L\), the two points approach different heights, so no single height fits both sides.
Press Make R equal L. The bar disappears, so one height fits both sides.
The two-sided limit is one height that both sides approach, so it exists only when \(R = L\), and then it is that common height.
Notation
A limit with no plus or minus sign above \(c\) is a two-sided limit, one number that \(f(x)\) approaches from both sides.
\[\lim_{x \to c} f(x)\]The two-sided limit exists when both one-sided limits exist and are equal. The limit is then their common value.
\[\begin{aligned} \lim_{x \to c^-} f(x) &= \cy{L} \\ \lim_{x \to c^+} f(x) &= \cy{L} \\ \Longrightarrow\ \lim_{x \to c} f(x) &= \cy{L} \end{aligned}\]The value of \(f\) at \(c\) is not part of this comparison.
If the one-sided limits are different, no single number is approached from both sides, so the two-sided limit does not exist.
\[\begin{aligned} &\cy{L} \neq \cy{R} \\ \Longrightarrow\ &\lim_{x \to c} f(x) \text{ does not exist} \end{aligned}\]Both one-sided limits must exist before you compare them. If one side has no limit, the two-sided limit does not exist.
When it fails
When the gap between L and R is 0, the sides agree and the limit exists. Any gap, however small, means that no single height fits both sides.
Move the points closer to the break. The gap stays the same size, so the two sides never agree.
Example 1: the sides meet
Let \(f(x) = x^2 - 1\) for \(x < 2\), let \(f(2) = 0\), and let \(f(x) = 2x - 1\) for \(x > 2\). Does \(\lim_{x \to 2} f(x)\) exist?
From the left, use \(x^2 - 1\). At \(x = 1.99\) the height is 2.9601, and the heights approach \(2^2 - 1 = 3\).
\[\lim_{x \to 2^-} f(x) = \cy{3}\]From the right, use \(2x - 1\). At \(x = 2.01\) the height is 3.02, and the heights approach \(2(2) - 1 = 3\).
\[\lim_{x \to 2^+} f(x) = \cy{3}\]The two sides agree, so the two-sided limit exists and equals 3.
\[\lim_{x \to 2} f(x) = \cy{3}\]The filled dot at \((2, 0)\) is the value \(f(2)\), and the comparison uses only the two sides, so the dot does not change the limit.
Example 2: the sides differ
Let \(g(x) = x + 2\) for \(x < -1\) and \(g(x) = 3 - x^2\) for \(x > -1\). Does \(\lim_{x \to -1} g(x)\) exist?
From the left, use \(x + 2\). The heights approach \(-1 + 2 = 1\).
\[\lim_{x \to -1^-} g(x) = \cy{1}\]From the right, use \(3 - x^2\). The heights approach \(3 - (-1)^2 = 2\).
\[\lim_{x \to -1^+} g(x) = \cy{2}\]A tempting wrong move: Both one-sided limits exist, so the limit as \(x\) approaches \(-1\) exists.
Both limits exist, but a limit needs one number that both sides approach. The red bar marks the error: the left side approaches 1 and the right side approaches 2, so no single height fits both sides.
Because \(1 \neq 2\), the sides disagree, so the two-sided limit does not exist.
\[\lim_{x \to -1} g(x) \text{ does not exist}\]
Practice
Let \(k(x) = 7 - x\) for \(x < 3\) and \(k(x) = x^2 - 5\) for \(x > 3\). Does \(\lim_{x \to 3} k(x)\) exist? If it does, find it.
Find the limit from each side. Compare the two heights on the graph.
Solution
From the left, the heights approach \(7 - 3 = 4\).
\[\lim_{x \to 3^-} k(x) = \cy{4}\]From the right, the heights approach \(3^2 - 5 = 4\).
\[\lim_{x \to 3^+} k(x) = \cy{4}\]The sides agree, so the limit exists.
\[\lim_{x \to 3} k(x) = \cy{4}\]
Questions
Use the graph for the first two questions. A filled dot shows the value of f at that input.
Does \(\lim_{x \to -2} f(x)\) exist?
- Yes, and it is \(3\).
- Yes, and it is \(1\).
- No, the limit does not exist.
Answer
Yes, and it is \(3\).
Does \(\lim_{x \to 1} f(x)\) exist?
- No, the limit does not exist.
- Yes, and it is \(0\).
- Yes, and it is \(1\).
Answer
No, the limit does not exist.
Let \(r(x) = 2x - 3\) for \(x < 4\), let \(r(4) = 0\), and let \(r(x) = x^2 - 11\) for \(x > 4\). What is \(\lim_{x \to 4} r(x)\)?
- \(5\)
- \(0\)
- The limit does not exist.
Answer
\(5\)
Summary
You find the limit from the left and the limit from the right, and if they are equal, the two-sided limit is that common number.
If they differ, the two-sided limit does not exist, because it must be one number that \(f(x)\) approaches from both sides.
Compare the two sides when a limit has no sign and the graph or the rule changes at the input.
This is one of four skills in topic 1.3 of Which & Why: AP Calculus.