AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/1/1-3)
Unit 1 · Topic 1.3
1.3 Estimating Limit Values from Graphs
A graph lets you see a limit: trace the curve toward x = c from the left and from the right and see what height you approach. The limit exists only if both sides head to the same height. Jumps, vertical asymptotes and wild oscillation are the usual reasons a limit fails.
Key terms
- one-sided limit
- left-hand limit
- right-hand limit
- does not exist (DNE)
- oscillation
How to read a limit off a graph
Put your finger on the curve to the left of x = c and slide it toward c. Note the y-value you approach: that is the left-hand limit. Do the same from the right for the right-hand limit.
Open circles (holes) and closed dots only tell you f(c). They don't change the limit. The limit is about where the curve is heading, not where the dot is.
Three ways a limit can fail
If the two sides don't agree on one height, or a side never settles, the limit does not exist. Here are the three patterns you'll see:
- Jump: the left side approaches one height and the right side another. Both one-sided limits exist, but they differ, so the two-sided limit does not exist.
- Unbounded behavior: the curve shoots up or down along a vertical asymptote. The outputs don't approach any number.
- Oscillation: the curve wiggles faster and faster between two heights as x approaches c and never settles. The classic example is sin(1/x) near x = 0, which swings between −1 and 1 infinitely often.
A checklist for graph questions
Graph questions are easy points if you're systematic. For each x-value the question names, work through these in order:
- Left-hand limit: where does the curve head as you come in from the left?
- Right-hand limit: where does it head from the right?
- Two-sided limit: it exists only if those two answers are the same number.
- Function value: where is the solid dot at that x? If there's only an open circle, f(c) is undefined.
- Compare: if the limit exists and equals the dot's height, the graph is unbroken there. That's continuity, coming in 1.11.
Estimating, not guessing
Graphs are only as precise as the drawing. If a graph's curve approaches a point that looks like (2, 3), say the limit appears to be 3. On the exam, graphs used for limits are drawn on grids so the values are clear, often at whole numbers or simple fractions.
If the graph shows a horizontal asymptote, you can also read limits at infinity off it: as x→∞, the curve's height approaches the asymptote's y-value. That idea returns in 1.15.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Reading limits at a jump and a hole
The graph of f has these features. As x approaches 1 from the left, the curve approaches the point (1, 2), which is an open circle. To the right of 1, the curve starts at a closed dot at (1, 4) and continues from there. At x = 3, the curve passes smoothly through an open circle at (3, 1), and there is a separate closed dot at (3, 5). Find lim (x→1⁻) f(x), lim (x→1⁺) f(x), lim (x→1) f(x), f(1), lim (x→3) f(x) and f(3).
Show the solutionHide the solution
- Step 1: Left of 1, the curve heads to height 2, so lim (x→1⁻) f(x) = 2.
- Step 2: Right of 1, the curve heads to height 4, so lim (x→1⁺) f(x) = 4.
- Step 3: The one-sided limits differ (2 ≠ 4), so lim (x→1) f(x) does not exist.
- Step 4: The closed dot at x = 1 is at height 4, so f(1) = 4.
- Step 5: At x = 3, both sides of the curve head into the hole at height 1, so lim (x→3) f(x) = 1.
- Step 6: The closed dot sits at height 5, so f(3) = 5. The dot doesn't affect the limit.
Answer: lim (x→1⁻) f(x) = 2, lim (x→1⁺) f(x) = 4, lim (x→1) f(x) does not exist, f(1) = 4, lim (x→3) f(x) = 1, f(3) = 5.
- Example 2
Oscillation near a point
The graph of y = sin(1/x) wiggles between y = −1 and y = 1, and the wiggles get infinitely tight as x approaches 0 from either side. Does lim (x→0) sin(1/x) exist?
Show the solutionHide the solution
- Step 1: As x gets close to 0, 1/x becomes huge, and sine of a huge input keeps cycling through every value from −1 to 1.
- Step 2: So in any tiny interval around 0, the outputs hit both 1 and −1 over and over. They never settle near a single number.
- Step 3: A limit requires the outputs to get and stay close to one value. That never happens here.
Answer: The limit does not exist because the function oscillates between −1 and 1 as x approaches 0.
- Example 3
Trap: a hole the screen doesn't show
On a graphing calculator, the graph of f(x) = (x² − 4)/(x − 2) looks like the straight line y = x + 2 with no gap. A student concludes that f(2) = 4. Is that right?
Show the solutionHide the solution
- Step 1: Plug in x = 2: (4 − 4)/(2 − 2) = 0/0, so f(2) is undefined. The graph really has a hole at (2, 4).
- Step 2: The calculator plots a finite number of points. A single missing point is too small to show at normal scale, so the screen hides it.
- Step 3: The limit is still 4: for x ≠ 2, f(x) = x + 2, which approaches 4.
Answer: No. lim (x→2) f(x) = 4, but f(2) does not exist. Graphs can miss behavior because of scale, so check the formula.
Common mistakes
- Reading the closed dot as the limit. The dot is f(c); the limit is where the curve heads.
- Saying a limit is the y-value of a hole only on one side. Check that the curve on both sides goes into the same hole.
- Calling oscillation “the limit is between −1 and 1.” If the outputs never settle, the limit simply does not exist.
On the exam
- A common multiple-choice question shows one graph and asks which of several limit statements is true. Check left limits, right limits, two-sided limits and function values separately.
- Graphs of piecewise functions show up in free response too, often to set up continuity or composite limit questions.
Connected topics
Videos
Check yourself
4 questions on 1.3 Estimating Limit Values from Graphs. Pick an answer to see if you got it, and why.
Which of the following best describes lim (x→0) sin(1/x) ?
What is lim (x→3) |x − 3|/(x − 3) ?
The function f is defined for all x in the closed interval [−3, 5] except x = 4. Its graph is described below.
On [−3, 0), the graph is a line segment from the point (−3, 0) to an open circle at (0, 2). The point (0, −1) is on the graph, so f(0) = −1.
On (0, 2], the graph is a line segment from an open circle at (0, 2) to the closed point (2, 4).
On (2, 4), the graph starts at an open circle at (2, 1) and rises, increasing without bound as x approaches 4 from the left.
On (4, 5], f(x) decreases without bound as x approaches 4 from the right, and the graph rises from there to the closed point (5, 0).
Described function
What is lim (x→0) f(x) ?
For which values of c in the open interval (−3, 5) does lim (x→c) f(x) fail to exist?
0 of 4 answered