AP Calculus BC 1.6 Determining Limits Using Algebraic Manipulation Study Notes - New Syllabus
AP Calculus BC 1.6 Determining Limits Using Algebraic Manipulation Study Notes- New syllabus
AP Calculus BC 1.6 Determining Limits Using Algebraic Manipulation 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:
- Limits Using Algebraic Manipulation
Determining Limits Using Algebraic Manipulation
Determining Limits Using Algebraic Manipulation
Sometimes direct substitution in a limit gives an indeterminate form like \( \frac{0}{0} \) or \( \infty – \infty \). In such cases, algebraic manipulation can be used to rewrite the expression into an equivalent form that allows evaluation of the limit.
Common Techniques:
- Factoring: Factor and cancel common terms.
- Rationalizing: Multiply by a conjugate when square roots are involved.
- Combining Fractions: Use common denominators to simplify differences of fractions.
- Simplifying Complex Fractions: Multiply numerator and denominator by a common factor.
Example (Factoring):
Find \( \lim_{x \to 3} \frac{x^2 – 9}{x – 3} \).
▶️ Answer/Explanation
Direct substitution gives \( \frac{0}{0} \), an indeterminate form.
Factor numerator: \( x^2 – 9 = (x – 3)(x + 3) \).
Simplify: \( \frac{(x – 3)(x + 3)}{x – 3} = x + 3 \) for \( x \neq 3 \).
Take the limit: \( \lim_{x \to 3} (x + 3) = 6 \).
Answer: \( 6 \).
Example (Rationalizing):
Compute \( \lim_{x \to 4} \frac{\sqrt{x} – 2}{x – 4} \).
▶️ Answer/Explanation
Direct substitution: \( \frac{2 – 2}{4 – 4} = \frac{0}{0} \).
Multiply numerator and denominator by conjugate \( \sqrt{x} + 2 \):
\( \frac{\sqrt{x} – 2}{x – 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x – 4}{(x – 4)(\sqrt{x} + 2)} \).
Simplify: \( \frac{1}{\sqrt{x} + 2} \).
Now take the limit: \( \frac{1}{2 + 2} = \frac{1}{4} \).
Answer: \( \frac{1}{4} \).
Example (Combining Fractions):
Evaluate \( \lim_{x \to 0} \frac{1}{x} – \frac{1}{x + 1} \).
▶️ Answer/Explanation
Combine into a single fraction:
\( \frac{(x + 1) – x}{x(x + 1)} = \frac{1}{x(x + 1)} \).
As \( x \to 0 \), the expression blows up to \( \infty \) (unbounded), so the two-sided limit does not exist.
But one-sided limits: As \( x \to 0^+ \), limit = \( +\infty \); as \( x \to 0^- \), limit = \( -\infty \).
Example (Simplifying Complex Fractions):
Compute \( \lim_{h \to 0} \frac{\frac{1}{x + h} – \frac{1}{x}}{h} \).
▶️ Answer/Explanation
Combine numerator:
\( \frac{1}{x + h} – \frac{1}{x} = \frac{x – (x + h)}{x(x + h)} = \frac{-h}{x(x + h)} \).
So expression becomes \( \frac{\frac{-h}{x(x + h)}}{h} = \frac{-1}{x(x + h)} \).
Take limit as \( h \to 0 \): \( \frac{-1}{x^2} \).
Answer: \( -\frac{1}{x^2} \).
Special Trigonometric Limits
Two important trigonometric limits frequently appear in Calculus. These limits form the basis of derivative formulas for trigonometric functions:
- \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
- \( \lim_{x \to 0} \frac{1 – \cos x}{x} = 0 \)
- \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \))
Key Notes:
- These limits are evaluated with \( x \) measured in radians.
- They are often proved using the Squeeze Theorem or geometry of the unit circle.
- They are useful for simplifying indeterminate forms involving trigonometric functions.
Example :
Evaluate \( \lim_{x \to 0} \frac{\sin(5x)}{x} \).
▶️ Answer/Explanation
Rewrite: \( \frac{\sin(5x)}{x} = \frac{\sin(5x)}{5x} \cdot 5 \).
As \( x \to 0 \), \( 5x \to 0 \), so \( \lim_{x \to 0} \frac{\sin(5x)}{5x} = 1 \).
Therefore, the limit = \( 1 \cdot 5 = 5 \).
Answer: \( 5 \).
Example :
Compute \( \lim_{x \to 0} \frac{1 – \cos(2x)}{x^2} \).
▶️ Answer/Explanation
Use the identity: \( 1 – \cos(2x) = 2\sin^2(x) \).
The expression becomes \( \frac{2\sin^2(x)}{x^2} = 2\left( \frac{\sin x}{x} \right)^2 \).
As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \).
So, the limit = \( 2(1)^2 = 2 \).
Answer: \( 2 \).
Example :
Find \( \lim_{x \to 0} \frac{\tan(3x)}{x} \).
▶️ Answer/Explanation
Rewrite: \( \frac{\tan(3x)}{x} = \frac{\tan(3x)}{3x} \cdot 3 \).
As \( x \to 0 \), \( \frac{\tan(3x)}{3x} \to 1 \) (since \( \lim_{u \to 0} \frac{\tan u}{u} = 1 \)).
So, the limit = \( 1 \cdot 3 = 3 \).
Answer: \( 3 \).