Unit 1 · Topic 1.3 Beta

Reading Limits from a Graph

Read a graph's limit one side at a time.

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

Graphs on the exam are rarely smooth curves. They come with holes, jumps, filled and open dots, and asymptotes, and most limit questions are about exactly those places. Reading a graph's limits is a skill of looking at each side separately. Once you can do that, every limit question on a graph takes seconds.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does limx→2f(x)=6\lim_{x \to 2} f(x) = 6 mean?

  1. As xx approaches 2, f(x)f(x) approaches 6
  2. f(2)=6f(2) = 6
  3. As xx approaches 6, f(x)f(x) approaches 2
  4. f(6)=2f(6) = 2
Show the answer

Input heads to 2, outputs head to 6; the value f(2)f(2) is not involved.

  • Correct: As xx approaches 2, f(x)f(x) approaches 6:
  • f(2)=6f(2) = 6:
  • As xx approaches 6, f(x)f(x) approaches 2:
  • f(6)=2f(6) = 2:

2. A graph has a filled dot at (3,1)(3, 1) and an open circle at (3,4)(3, 4). What is f(3)f(3)?

  1. 11
  2. 44
  3. not defined
  4. 33
Show the answer

The filled dot is on the graph.

  • Correct: 11:
  • 44:
  • not defined:
  • 33:

Part 4 · See it

See it first

Four small graphs near x = 1: a curve with an open circle at (1, 2) and a dot at (1, 3), limit 2; a jump from height 1 on the left to height 3 on the right, no limit; a graph rising without bound on both sides of x = 1, no limit; a graph oscillating faster and faster between 1 and 3 near x = 1, no limit.
Four behaviors as xx approaches 1: a hole (limit 2), a jump (no limit), unbounded outputs (no limit), oscillation (no limit). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Slide xx toward aa from the left onlythe heights give the left-hand limit, limx→a−f(x)\lim_{x \to a^-} f(x)
  2. Slide xx toward aa from the right onlythe heights give the right-hand limit, limx→a+f(x)\lim_{x \to a^+} f(x)
  3. Both one-sided limits exist and are equalthe two-sided limit exists and equals that number
  4. The sides disagree, the outputs are unbounded, or they oscillatethe limit does not exist

Part 6 · Key ideas

Key ideas

  • limx→a−f(x)\lim_{x \to a^-} f(x) uses only x<ax < a; limx→a+f(x)\lim_{x \to a^+} f(x) uses only x>ax > a. Follow the graph's heights on that side to the edge at x=ax = a.
  • limx→af(x)=L\lim_{x \to a} f(x) = L exactly when both one-sided limits equal LL.
  • A limit fails to exist when the one-sided limits differ (a jump), when the outputs grow without bound, or when they oscillate.
  • Filled and open dots tell you f(a)f(a); they never decide the limit.
  • A justification names both one-sided limits: "From the left f(x)→3f(x) \to 3; from the right f(x)→1f(x) \to 1; since 3≠13 \ne 1, the limit does not exist."

Part 7 · Misconception

A common mistake

The wrong idea: If f(a)f(a) is not defined, then limx→af(x)\lim_{x \to a} f(x) does not exist.

What actually happens: A missing value never decides a limit. A hole has a limit (the height it sits at) and no value. Decide existence only from the two one-sided limits.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of ff

The graph of ff for −4≤x≤4-4 \le x \le 4 has three pieces. A line rises from (−4,1)(-4, 1) to an open circle at (−2,3)(-2, 3); a filled dot sits at (−2,1)(-2, 1). A curve leaves the open circle at (−2,3)(-2, 3) and falls to the filled point (1,−1.5)(1, -1.5). A line starts at an open circle at (1,0)(1, 0) and rises to (4,3)(4, 3).
−4−3−2−11234−2−11234xyf

1. Reading each side of the graph, what is limx→−2f(x)\lim_{x \to -2} f(x)?

  1. 33
  2. 11
  3. −2-2
  4. The limit does not exist.
Show the answer

From the left, the line heads to height 3; from the right, the curve also heads to height 3. Both one-sided limits are 3, so the limit is 3. The dot at (−2,1)(-2, 1) is f(−2)f(-2), which does not affect the limit.

  • Correct: 33: Left and right both head to 3.
  • 11: 1 is f(−2)f(-2), the filled dot. A limit does not use the value at the point.
  • −2-2: −2-2 is the input being approached.
  • The limit does not exist.: Both sides head to the same height, 3, so the limit exists.

2. What is limx→1+f(x)\lim_{x \to 1^+} f(x)?

  1. 00
  2. −1.5-1.5
  3. 11
  4. The limit does not exist.
Show the answer

For xx just to the right of 1, the graph is the rising line, whose heights head to 0, the open circle at (1,0)(1, 0).

  • Correct: 00: From the right, the line heads to the open circle at height 0.
  • −1.5-1.5: −1.5-1.5 is where the curve on the left ends, so it is the limit from the left.
  • 11: 1 is the xx-value, not a height.
  • The limit does not exist.: A one-sided limit only needs one side to settle, and the right side heads to 0.

3. Which statement about limx→1f(x)\lim_{x \to 1} f(x) is correct?

  1. It does not exist, because limx→1−f(x)\lim_{x \to 1^-} f(x)=−1.5= -1.5 and limx→1+f(x)=0\lim_{x \to 1^+} f(x) = 0.
  2. It equals −1.5-1.5, because f(1)=−1.5f(1) = -1.5.
  3. It equals −0.75-0.75, halfway between the two sides.
  4. It does not exist, because f(1)f(1) is not defined.
Show the answer

A two-sided limit needs both one-sided limits to be equal. From the left the heights head to −1.5-1.5; from the right, to 0. They differ, so the limit does not exist.

  • Correct: It does not exist, because limx→1−f(x)\lim_{x \to 1^-} f(x)=−1.5= -1.5 and limx→1+f(x)=0\lim_{x \to 1^+} f(x) = 0.: Names both one-sided limits and why they rule out a limit.
  • It equals −1.5-1.5, because f(1)=−1.5f(1) = -1.5.: f(1)=−1.5f(1) = -1.5 is a value; it does not make the right side agree.
  • It equals −0.75-0.75, halfway between the two sides.: There is no averaging: when the sides disagree there is no limit.
  • It does not exist, because f(1)f(1) is not defined.: f(1)f(1) is defined: the filled point (1,−1.5)(1, -1.5). The reason is the unequal sides.

4. Which statement about ff is true?

  1. limx→1−f(x)\lim_{x \to 1^-} f(x)=f(1)= f(1)
  2. limx→1+f(x)\lim_{x \to 1^+} f(x)=f(1)= f(1)
  3. limx→−2f(x)\lim_{x \to -2} f(x)=f(−2)= f(-2)
  4. limx→−2−f(x)=1\lim_{x \to -2^-} f(x) = 1
Show the answer

From the left of 1, the curve heads to −1.5-1.5, and the filled point at the end of that curve says f(1)=−1.5f(1) = -1.5. They match.

  • Correct: limx→1−f(x)\lim_{x \to 1^-} f(x)=f(1)= f(1): Both equal −1.5-1.5.
  • limx→1+f(x)\lim_{x \to 1^+} f(x)=f(1)= f(1): From the right the heights head to 0, but f(1)=−1.5f(1) = -1.5.
  • limx→−2f(x)\lim_{x \to -2} f(x)=f(−2)= f(-2): The limit at −2-2 is 3, but f(−2)=1f(-2) = 1.
  • limx→−2−f(x)=1\lim_{x \to -2^-} f(x) = 1: From the left of −2-2, the line heads to 3, not to the dot at 1.

5. Find limx→−2−f(x)+f(−2)\lim_{x \to -2^-} f(x) + f(-2).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From the left, the line's heights head to 3. The value f(−2)f(-2) is the filled dot, 1. Sum: 3+1=43 + 1 = 4.

  • Answer: 4

6. Near x=0x = 0, the graph of a function qq wiggles up and down between heights −1-1 and 11 infinitely many times, faster and faster as xx gets closer to 0. What is limx→0q(x)\lim_{x \to 0} q(x)?

  1. It does not exist, because the outputs oscillate and settle on no single number.
  2. 00, the height halfway between −1-1 and 11.
  3. 11, the highest height the graph reaches.
  4. It does not exist, because the outputs grow without bound.
Show the answer

However close xx gets to 0, the outputs keep taking every value from −1-1 to 1. They never settle near one number, so there is no limit.

  • Correct: It does not exist, because the outputs oscillate and settle on no single number.: Oscillating outputs settle on nothing.
  • 00, the height halfway between −1-1 and 11.: The outputs pass through 0 but never stay close to it.
  • 11, the highest height the graph reaches.: The graph keeps coming back down from 1.
  • It does not exist, because the outputs grow without bound.: The outputs stay between −1-1 and 1; they are bounded. The cause is oscillation.

7. Which condition guarantees that limx→af(x)=5\lim_{x \to a} f(x) = 5?

  1. limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5 and limx→a+f(x)=5\lim_{x \to a^+} f(x) = 5
  2. f(a)=5f(a) = 5
  3. limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5
  4. f(a)=5f(a) = 5 and limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5
Show the answer

A two-sided limit exists and equals LL exactly when both one-sided limits exist and equal LL.

  • Correct: limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5 and limx→a+f(x)=5\lim_{x \to a^+} f(x) = 5: Both sides equal 5: that is the definition's test.
  • f(a)=5f(a) = 5: The value at aa says nothing about nearby outputs.
  • limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5: One side is not enough; the right side could head elsewhere.
  • f(a)=5f(a) = 5 and limx→a−f(x)=5\lim_{x \to a^-} f(x) = 5: The right side is still unknown, and the value does not help.

Part 9 · Summary

Summary

Read a graph's limit one side at a time. The left-hand and right-hand limits are the heights the graph heads to from each side. If they exist and agree, that is the limit; if they differ, or the outputs are unbounded or oscillate, the limit does not exist. Dots give the value f(a)f(a), which never affects the limit.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections