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{x \to 0} \frac{x}{x} + \lim{x \to 0} \frac{\sin x}{x} .
Solve Term 1: For \lim{x \to 0} \frac{x}{x} , the x values cancel out (since x \neq 0 as x approaches 0). This simplifies to \lim{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