AP Calculus BC 1.5 Determining Limits Using Algebraic Properties of Limits Study Notes - New Syllabus
AP Calculus BC 1.5 Determining Limits Using Algebraic Properties of Limits Study Notes- New syllabus
AP Calculus BC 1.5 Determining Limits Using Algebraic Properties of Limits Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Determining Limits Using Algebraic Properties of Limits
Key Concepts:
- Determining Limits Using Algebraic Properties of Limits
Determining Limits Using Algebraic Properties of Limits
Determining Limits Using Algebraic Properties of Limits
The value of a limit can often be determined using the limit laws (algebraic properties). These properties allow us to break down complex expressions into simpler components.
Key Limit Laws:
| Property | Description |
|---|---|
| \( \lim_{x \to a} [f(x) + g(x)] = L + M \) | Sum of limits = Limit of sum |
| \( \lim_{x \to a} [f(x) – g(x)] = L – M \) | Difference of limits = Limit of difference |
| \( \lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M \) | Product of limits = Limit of product |
| \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, M \neq 0 \) | Quotient of limits = Limit of quotient |
| \( \lim_{x \to a} c \cdot f(x) = c \cdot L \) | Constant multiple law |
| \( \lim_{x \to a} [f(g(x))] = f\big(\lim_{x \to a} g(x)\big) \) | Limit of a composite function (if continuous) |
Conditions:
- These laws apply when individual limits exist and, in the quotient law, the denominator’s limit is not zero.
- Composite function law applies when the outer function is continuous at the inner limit.
Example :
Compute \( \lim_{x \to 3} (x^2 + 5x – 7) \).
▶️ Answer/Explanation
Apply sum and constant laws:
\( \lim_{x \to 3} x^2 = 3^2 = 9 \)
\( \lim_{x \to 3} 5x = 5(3) = 15 \)
\( \lim_{x \to 3} (-7) = -7 \)
Combine: \( 9 + 15 – 7 = 17 \)
Answer: \( 17 \)
Example :
Find \( \lim_{x \to 2} \frac{x^2 – 4}{x – 2} \).
▶️ Answer/Explanation
Direct substitution gives \( \frac{4 – 4}{2 – 2} = \frac{0}{0} \), an indeterminate form.
Factor numerator: \( x^2 – 4 = (x – 2)(x + 2) \).
Simplify: \( \frac{(x – 2)(x + 2)}{x – 2} = x + 2 \) for \( x \neq 2 \).
Now take limit: \( \lim_{x \to 2} (x + 2) = 4 \).
Answer: \( 4 \)
Example :
Evaluate \( \lim_{x \to 0} (\sin x + x^2) \).
▶️ Answer/Explanation
Use sum law:
\( \lim_{x \to 0} \sin x = 0 \) and \( \lim_{x \to 0} x^2 = 0 \).
So, \( \lim_{x \to 0} (\sin x + x^2) = 0 + 0 = 0 \).
Example :
Compute \( \lim_{x \to -2} [7x^3] \).
▶️ Answer/Explanation
Apply constant multiple law: \( 7 \cdot \lim_{x \to -2} (x^3) \).
\( x^3 \) at \( x = -2 \) is \( (-2)^3 = -8 \).
So, \( 7 \cdot (-8) = -56 \).
Answer: \( -56 \).
Example:
Find \( \lim_{x \to 1} \frac{x^2 + 3x – 4}{x – 1} \).
▶️ Answer/Explanation
Direct substitution gives \( \frac{(1)^2 + 3(1) – 4}{1 – 1} = \frac{0}{0} \), so factor numerator.
\( x^2 + 3x – 4 = (x – 1)(x + 4) \).
Simplify: \( \frac{(x – 1)(x + 4)}{x – 1} = x + 4 \) for \( x \neq 1 \).
Take the limit: \( \lim_{x \to 1} (x + 4) = 5 \).
Answer: \( 5 \).
Example:
Evaluate \( \lim_{x \to 0^+} \sqrt{9 + 4x} \).
▶️ Answer/Explanation
The outer function \( \sqrt{\phantom{x}} \) is continuous, so we can substitute the inner limit.
\( \lim_{x \to 0^+} (9 + 4x) = 9 \).
Then, \( \sqrt{9} = 3 \).
Answer: \( 3 \).
Example :
Compute \( \lim_{x \to 2} \frac{5x^2 – 3}{x + 1} \).
▶️ Answer/Explanation
Apply quotient law:
\( \lim_{x \to 2} (5x^2 – 3) = 5(2^2) – 3 = 20 – 3 = 17 \).
\( \lim_{x \to 2} (x + 1) = 2 + 1 = 3 \).
So, \( \frac{17}{3} \).
Answer: \( \frac{17}{3} \approx 5.67 \).