Recognize a jump
A fee that starts after 2 hours
The graph shows the cost \(C(t)\) in dollars of parking for \(t\) hours. The meter charges 3 dollars per hour, and after 2 hours it adds a fee of 4 dollars.
Move \(t\) toward 2 from the left. The cost heads to 6 dollars. Move \(t\) toward 2 from the right. The cost heads to 10 dollars.
The two sides head to different heights, so the graph jumps by 4 dollars at \(t=2\), and you must lift the pen to draw it. The filled dot shows that \(C(2)=6\).
Two sides, two heights
The graph now shows a general break at an input \(c\). The left piece heads to a height \(\cy{L_1}\), and the right piece heads to a height \(\cy{L_2}\).
These heights are the one-sided limits. The left-hand limit uses only inputs below \(c\), and the right-hand limit uses only inputs above \(c\).
If \(L_1\) and \(L_2\) are different, the graph has a jump at \(c\). Move the slider. The bar between the rings shows the jump, green when the right piece is higher and red when it is lower.
No single height fits both sides, so the two-sided limit does not exist. A value at \(c\) cannot fill the gap, because a value is one height and the gap has two ends.
Notation
The left-hand limit is the height that the left piece approaches. It uses only inputs less than \(c\).
\[\lim_{x\to c^-} f(x)=\cy{L_1}\]The right-hand limit uses only inputs greater than \(c\).
\[\lim_{x\to c^+} f(x)=\cy{L_2}\]If both limits are finite and they are different, \(f\) has a jump at \(c\).
\[L_1\neq L_2\]The size of the jump is the right height minus the left height. It is positive when the graph steps up and negative when it steps down.
\[\text{jump}=L_2-L_1\]The two-sided limit does not exist, because no single number is approached from both sides.
\[\lim_{x\to c} f(x)\ \text{does not exist}\]
When it fails
Move the right piece down with the slider. The two open rings move closer, and when the heights are equal, both sides head to the same number.
A jump needs two different heights. With equal heights the two-sided limit exists, and the break is only a hole.
Example 1: a step down
Let \(j(x)=x^2\) for \(x<3\), and \(j(x)=2x+1\) for \(x\geq 3\). Describe the break at \(x=3\).
From the left, \(x^2\) heads to 9.
\[\lim_{x\to 3^-} j(x)=\cy{9}\]From the right, \(2x+1\) heads to 7.
\[\lim_{x\to 3^+} j(x)=\cy{7}\]Both limits are finite and they are different, so \(j\) has a jump at 3. The graph steps down by 2, so the size of the jump is negative, and the red bar marks the drop.
\[7-9=\cr{-2}\]The filled dot shows \(j(3)=7\) on the right piece, so the gap up to 9 remains.
Example 2: a value that fills one end
Let \(w(x)=8-x^2\) for \(x<2\), and \(w(x)=3x-1\) for \(x>2\). Can any value \(w(2)\) make \(w\) continuous at 2?
From the left, \(8-x^2\) heads to 4. From the right, \(3x-1\) heads to 5.
\[\lim_{x\to 2^-} w(x)=\cy{4},\quad \lim_{x\to 2^+} w(x)=\cy{5}\]A tempting wrong move: Set \(w(2)=4\). Then the dot fills the gap, and \(w\) is continuous at 2.
\[w(2)=\cr{4}\]The red dot at height 4 meets the left piece only. The right piece still starts at height 5, so the pen must still lift by 1. The green bar shows that rise.
Continuity at 2 needs both one-sided limits to equal \(w(2)\). One dot cannot be at height 4 and at height 5, so no value of \(w(2)\) makes \(w\) continuous at 2.
\[5-4=1\]
Practice
Let \(s(x)=\dfrac{2|x+3|}{x+3}\) for \(x\neq -3\). Find both one-sided limits at \(-3\) and the size of the jump.
Above \(-3\), the number \(x+3\) is positive, so \(|x+3|\) equals \(x+3\). Below \(-3\), \(|x+3|\) equals \(-(x+3)\).
Solution
For \(x>-3\), the fraction is \(\frac{2(x+3)}{x+3}\), which equals 2. The right-hand limit is 2.
\[\lim_{x\to -3^+} s(x)=\cy{2}\]For \(x<-3\), the fraction is \(\frac{-2(x+3)}{x+3}\), which equals \(-2\). The left-hand limit is \(-2\).
\[\lim_{x\to -3^-} s(x)=\cy{-2}\]The limits are different, so \(s\) has a jump at \(-3\). Its size is the right-hand limit minus the left-hand limit, and the green bar shows this rise.
\[2-(-2)=4\]
Questions
The graph shows a function S made of flat steps. Each step starts with a filled dot, and the first three steps end with an open ring.
What is the left-hand limit of \(S\) as \(x\) approaches 3?
- 5
- 7
- 2
- It does not exist
Answer
5
What is the size of the jump of \(S\) at \(x=2\)?
- 1
- 5
- 4
- 9
Answer
1
Let \(h(x)=3x^2-2\) for \(x<1\), and \(h(x)=4x-1\) for \(x\geq 1\). What is the size of the jump of \(h\) at 1?
- 2
- 1
- 3
- \(-2\)
Answer
2
Summary
You find the left-hand limit and the right-hand limit, here 6 and 10, and you compare them.
This works because a one-sided limit uses only the graph on that side, so two different heights mean that the pieces do not meet.
Use it where a graph or formula changes, to tell a jump from a hole and to find its size.
This is one of five skills in topic 1.10 of Which & Why: AP Calculus.