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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

AP Calculus BC-Concise Summary Notes- All Topics

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: \( 3 \).

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