Sept 19th - Limits : Trig Substitution

Homework and Quiz Information

• Upcoming homework quizzes will be based on material from section 2.1.

• Problems will vary for different classes and individual students to ensure fairness and prevent sharing direct answers.

• The specific types of problems (e.g., graphs vs. algebraic limits) on the quiz are random.

• Occasionally, if certain problems are deemed most effective for AP test preparation (e.g., graph problems), similar types might appear across classes, but the specific numerical problems will differ.

Lost and Found

• A check for a lost water bottle or bracelet was made, as items had been left in the classroom for approximately a week.

Limits Using Algebra and Trigonometric Identities

• Key Principle: Trigonometric identities like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ only apply when the limit is approaching $0$. If the limit approaches a different value, these specific identities cannot be used directly.

• Example 1: Combining Algebra and Trig Identities

  • Problem: Calculate $\lim_{x \to 0} \frac{x + \sin x}{x}$.

  • Initial Check: Direct substitution of $x=0$ yields $0/0$, an indeterminate form.

  • Strategy: Separate the expression algebraically using limit laws established after Theorem 1.

    • $\lim_{x \to 0} \left( \frac{x}{x} + \frac{\sin x}{x} \right)$

    • This can be written as $\lim<em>{x \to 0} \frac{x}{x} + \lim</em>{x \to 0} \frac{\sin x}{x}$.

  • Solve Term 1: For $\lim<em>{x \to 0} \frac{x}{x}$, the $x$ values cancel out (since $x \neq 0$ as $x$ approaches $0$). This simplifies to $\lim</em>{x \to 0} 1 = 1$.

  • Solve Term 2: For $\lim_{x \to 0} \frac{\sin x}{x}$, this is a direct application of the known trigonometric identity, which equals $1$.

  • Final Answer: Adding the results, $1 + 1 = 2$.

  • Interpretation: If one were to graph $y = \frac{x + \sin x}{x}$, the $y$-values would approach $2$ as $x$ approaches $0$.

• Example 2: Algebraic Simplification (Difference of Squares) for Trig Functions

  • Problem: Calculate $\lim_{x \to \pi/4} \frac{\sin^2 x - \cos^2 x}{\sin x - \cos x}$.

  • Initial Check: Direct substitution of $x = \pi/4$ leads to $(\frac{\sqrt{2}}{2})^2 - (\frac{\sqrt{2}}{2})^2$ in the numerator ($1/2 - 1/2 = 0$) and $\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}$ in the denominator ($0$). This is an indeterminate form $0/0$.

  • Why trig identities (like $\frac{\sin x}{x}$) don't work here: The limit is approaching $\pi/4$, not $0$.

  • Strategy: Recognize the numerator as a difference of squares. Despite involving trigonometric functions, the algebraic property $a^2 - b^2 = (a-b)(a+b)$ still applies.

    • Factor the numerator: $(\sin x - \cos x)(\sin x + \cos x)$.

    • The expression becomes: $\lim_{x \to \pi/4} \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x}$.

  • Cancel Terms: The common factor $(\sin x - \cos x)$ cancels out (since $\sin x - \cos x \neq 0$ as $x$ approaches $\pi/4$).

  • New Limit: $\lim_{x \to \pi/4} (\sin x + \cos x)$.

  • Direct Substitution: Now, direct substitution is valid.

    • $\sin(\pi/4) + \cos(\pi/4) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2}$.

Trig Substitution: A "Get Out of Jail Free" Card

• Purpose: When a limit involving trig functions results in an indeterminate form (e.g., $0/0$) via direct substitution, and standard algebraic methods (like factoring) or basic trig identities (like $\lim_{x \to 0} \frac{\sin x}{x}$) don't apply (especially when not approaching $0$ or if the functions are of different types, e.g., linear and trig).

• Goal: To transform the limit into a form that does approach $0$, allowing the use of basic trig identities.

• Example 3: Trig Substitution with Sine Addition Formula

  • Problem: Calculate $\lim_{x \to \pi/2} \frac{(2x - \pi)}{\csc(2x)}$.

  • Initial Check: Direct substitution of $x = \pi/2$ yields $(2(\pi/2) - \pi) = 0$ in the numerator. The denominator is $\csc(2(\pi/2)) = \csc(\pi) = \frac{1}{\sin(\pi)} = \frac{1}{0}$ (undefined, but in the context of limits it leads to $0/0$ when considering the reciprocal). Indeterminate form $0/0$.

  • Why standard methods fail: This is a linear function over a trig function. They are from different