AP Calculus BC 1.9 Connecting Multiple Representations of Limits Study Notes - New Syllabus
AP Calculus BC 1.9 Connecting Multiple Representations of Limits Study Notes- New syllabus
AP Calculus BC 1.9 Connecting Multiple Representations of Limits Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
Reasoning with definitions, theorems, and properties can be used to justify claims about limits.
Key Concepts:
- Linking multiple representations of limits (graphical, numerical, analytical, and contextual)
Connecting Multiple Representations of Limits
Connecting Multiple Representations of Limits
Limits can be represented and interpreted in different forms:
- Analytical: Using algebraic expressions and limit notation.
- Graphical: Observing behavior of a function on a graph.
- Numerical (Table): Using a table of values approaching a point.
- Verbal: Describing behavior in words.
Key Skill: Be able to move between these representations to interpret and evaluate limits.
Steps for Connecting Representations:
- Start with the given representation (table, graph, or equation).
- Analyze the behavior as \( x \) approaches a value from both sides.
- Express the observation using correct limit notation.
- Verify consistency between numerical, graphical, and analytical approaches.
Example :
The table shows values of \( f(x) = \dfrac{\sin x}{x} \) as \( x \) approaches 0:
| \( x \) | -0.1 | -0.01 | -0.001 | 0.001 | 0.01 | 0.1 |
|---|---|---|---|---|---|---|
| \( f(x) \) | 0.9983 | 0.99998 | 1.0000 | 1.0000 | 0.99998 | 0.9983 |
▶️ Answer/Explanation
From the table, as \( x \to 0 \), \( f(x) \to 1 \).
Limit: \( \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \).
Example :
Use the table to estimate \( \lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} \).
| \( x \) | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| \( f(x) \) | 3.9 | 3.99 | 3.999 | 4.001 | 4.01 | 4.1 |
▶️ Answer/Explanation
As \( x \) approaches 2 from both sides, \( f(x) \) approaches 4.
Estimated limit: \( \lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} = 4 \).
(Later, we verify analytically: \( \dfrac{(x-2)(x+2)}{x-2} \to x+2 = 4 \)).
Example :
Let \( g \) be a function where \( \lim_{x \to 3} g(x) = 6 \). Which of the following could represent the function \( g \)?
a) \( g(x) = \begin{cases} \dfrac{x^2 – x – 6}{x – 3}, & x \neq 3 \\ 6, & x = 3 \end{cases} \)
b) Table: Two linear pieces
| \( x \) | 2.8 | 2.9 | 2.999 | 3.001 | 3.1 | 3.2 |
|---|---|---|---|---|---|---|
| \( g(x) \) | 6.2 | 6.01 | 6.001 | 4.999 | 4.9 | 4.8 |
c) Graph:
▶️ Answer/Explanation
Option (a): Check option (a): For \( x \neq 3 \), simplify \( \dfrac{x^2 – x – 6}{x – 3} \):
\[ x^2 – x – 6 = (x – 3)(x + 2), \text{ so } \dfrac{(x – 3)(x + 2)}{x – 3} = x + 2. \]
So \( \lim_{x \to 3} (x + 2) = 5 \), not 6 → option (a) is incorrect.
Option (b): Option (b): Table shows values approaching 6 from the left and near 5 from the right → does not satisfy \( \lim = 6 \).
Option (c): The graph shows that as \( x \) approaches 3 from both sides, \( g(x) \) approaches 6 (open circle at (3,6) and point (3,3) does not affect the limit).
$\lim_{x \to 3} g(x) = 6.$ So (c) is correct.
Correct answer:The graph represents the correct situation where the limit exists and equals 6, even though the function’s value at \( x = 3 \) is different (discontinuity).
Example :
If \( h \) is a piecewise function with two linear pieces such that \( \lim_{x \to 4} h(x) \) does not exist, which of the following could represent the function \( h \)?
a) \( h(x) = \begin{cases} \dfrac{1}{2}x + 3, & x < 4 \\ 13 – 2x, & x > 4 \end{cases} \)
b) Table:
| \( x \) | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| \( h(x) \) | -4 | -1 | 2 | 1 | \( \dfrac{16}{3} \) | \( \dfrac{17}{3} \) | \( \dfrac{18}{3} \) |
c) Graph:
▶️ Answer/Explanation
Option (a): Compute one-sided limits: \( \lim_{x \to 4^-} h(x) = \dfrac{1}{2}(4) + 3 = 5 \), \( \lim_{x \to 4^+} h(x) = 13 – 2(4) = 5 \). Both are equal, so the limit exists. No, the limit exists.
Option (b): From the table: As \( x \to 4^- \), values approach about 2. As \( x \to 4^+ \), values approach about \( \dfrac{16}{3} \approx 5.33 \). They are not equal, so the limit does NOT exist. Correct: Limit does not exist.
Option (c): The graph shows a jump at \( x = 4 \): Left-hand limit \( \approx 3.5 \), right-hand limit \( = 5 \). They differ, so the limit does NOT exist. Correct: Limit does not exist.
Example:
If \( f(x) = \begin{cases} \dfrac{(x-3)^2(x^2+1)}{|x-3|}, & x \neq 3 \\ 2, & x = 3 \end{cases} \), then \( \lim_{x \to 3} f(x) \) is:
(A) 0
(B) 2
(C) 10
(D) Nonexistent
▶️ Answer/Explanation
Simplify for \( x \neq 3 \):
\[ f(x) = \dfrac{(x-3)^2(x^2+1)}{|x-3|} = |x-3|(x^2+1). \]
As \( x \to 3 \), \( |x-3| \to 0 \) and \( x^2+1 \to 10 \), so the product \( \to 0 \).
Answer: (A) 0.