Compare the two sides
A curve with two parts
The curve has two parts. The left part is \(f(x) = x + 1\) for \(x < 2\), and the right part is \(f(x) = x + 3\) for \(x > 2\). Each part ends at an open ring at \(x = 2\).
One orange dot sits on each part. Drag the left dot toward 2, and the right dot moves toward 2 from the other side.
At a distance of 0.1 from 2, the left output is 2.9 and the right output is 5.1. At a distance of 0.01, the outputs are 2.99 and 5.01.
The left outputs approach 3, so the left-hand limit is \(\lim_{x \to 2^-} f(x) = 3\), where the small minus sign means that \(x\) stays less than 2.
The right outputs approach 5, so the right-hand limit is \(\lim_{x \to 2^+} f(x) = 5\), where the plus sign means that \(x\) stays greater than 2.
Because 3 and 5 differ, the outputs near 2 do not approach one number. The two-sided limit, which uses both sides at once, does not exist.
Two heights at one input
At the input \(c\), the left part of the curve approaches the height \(L_1\) and the right part approaches the height \(L_2\). The orange lines mark the two heights.
Drag the slider to move the right part up or down. When the two heights differ, the outputs from the two sides approach different numbers, so the limit does not exist.
When the two heights are equal, the two parts meet at one height, and that height is the two-sided limit.
The two-sided limit exists only when both one-sided limits exist and are equal.
Notation
The left-hand limit uses only inputs less than \(c\), so it is the height that the left part approaches.
\[\lim_{x \to c^-} f(x) = \cy{L_1}\]The right-hand limit uses only inputs greater than \(c\), so it is the height that the right part approaches.
\[\lim_{x \to c^+} f(x) = \cy{L_2}\]The two-sided limit exists when both one-sided limits exist and are equal. It then has that common value.
\[\begin{aligned} \lim_{x \to c} f(x) = L \quad &\text{exactly when} \\ \lim_{x \to c^-} f(x) = L \quad &\text{and} \\ \lim_{x \to c^+} f(x) = L \end{aligned}\]If a one-sided limit does not exist, or if the two differ, the two-sided limit does not exist.
When it fails
Drag the slider to raise the right part. Each one-sided limit still exists, but the two now differ by the size of the jump.
The two-sided limit then does not exist, even though both one-sided limits do.
Example 1: the formula changes and the limit exists
Find \(\lim_{x \to 3} h(x)\), where \(h(x) = 2x + 1\) for \(x < 3\) and \(h(x) = 10 - x\) for \(x > 3\).
The formula changes at 3, so find each side on its own. For \(x < 3\), the output \(2x + 1\) approaches \(2(3) + 1 = 7\), as the left dot shows.
\[\lim_{x \to 3^-} (2x + 1) = \cy{7}\]For \(x > 3\), the output \(10 - x\) approaches \(10 - 3 = 7\), as the right dot shows.
\[\lim_{x \to 3^+} (10 - x) = \cy{7}\]Both one-sided limits equal 7, so the two-sided limit exists and equals 7.
\[\lim_{x \to 3} h(x) = \cy{7}\]
Example 2: an absolute value
Find \(\lim_{x \to 3} g(x)\), where \(g(x) = \dfrac{|2x - 6|}{x - 3}\).
For \(x > 3\), the number \(2x - 6\) is positive, so \(|2x - 6| = 2x - 6 = 2(x - 3)\), and the quotient is 2. The curve is green where its outputs are positive and red where they are negative.
\[\lim_{x \to 3^+} \dfrac{2x - 6}{x - 3} = \cy{2}\]A tempting wrong move: The right side gives 2, so the limit is 2.
\[\lim_{x \to 3} g(x) = 2\]The red part of the curve, on the left of 3, lies at height −2, far from the red line at 2. A limit needs the outputs on both sides to approach the same number.
For \(x < 3\), the number \(2x - 6\) is negative, so \(|2x - 6| = -2(x - 3)\), and the quotient is \(-2\). The one-sided limits differ, so the limit does not exist.
\[\lim_{x \to 3^-} g(x) = \cr{-2}\]
Practice
Does \(\lim_{x \to 1} p(x)\) exist, where \(p(x) = x^2 + 2\) for \(x < 1\) and \(p(x) = 6 - 2x\) for \(x > 1\)?
Move the dots toward 1. Find the height that each part approaches. Then compare the two heights.
Solution
Near 1, \(x^2 + 2\) is near \(1^2 + 2 = 3\), so the left part approaches 3.
\[\lim_{x \to 1^-} (x^2 + 2) = \cy{3}\]Near 1, \(6 - 2x\) is near \(6 - 2(1) = 4\), so the right part approaches 4.
\[\lim_{x \to 1^+} (6 - 2x) = \cy{4}\]The one-sided limits differ, so the limit does not exist.
Questions
The graph shows a function \(q\) with a break at \(x = 1\) and another at \(x = 4\).
What is the right-hand limit of \(q\) at \(x = 1\)?
- 3
- 2
- It does not exist
Answer
3
Does the two-sided limit of \(q\) exist at \(x = 4\)?
- Yes, and it equals 0
- No, because the formula changes at 4
- No, because \(q(4)\) is not defined
Answer
Yes, and it equals 0
Find \(\lim_{x \to 5} f(x)\), where \(f(x) = 2x - 3\) for \(x < 5\) and \(f(x) = 12 - x\) for \(x > 5\).
- 7
- It does not exist, because the formula changes at 5
- 5
Answer
7
Summary
You find the left-hand limit and the right-hand limit separately, from the formula or the graph on each side.
The two-sided limit exists only when they are equal, because the outputs on both sides of the input must approach the same number.
Compare the two sides where a formula changes, where an absolute value appears, or where a graph breaks.
This is one of four skills in topic 1.2 of Which & Why: AP Calculus.