Sept 17th - Squeeze Theorem and Trig Limits

Announcements and Class Logistics

  • Title test to celebrate Hispanic heritage with participation points and a cord, and as a candidate for Spanish club executive board.
  • Homecoming dance tickets on October 4 sold starting Friday, September 19 in GoFan; tickets will not be sold at the door; buy before October 3 at midnight.
  • Powder puff team for girls: sign up in the avail sign up through the athletic office until October 23.
  • Bayport FFA new member sign up meeting on Wednesday, September 24 during part time; sign up in flex in Room AA114.
  • Travel Night: learn about four upcoming educational experiences abroad (Northern Italy, Galapagos Islands, Japan, or Iceland); Travel Night on Wednesday, September 24 at 6PM in Bayport’s Forum Room. See the Schoology post from this morning and sign up for more information.
  • If interested in Bayport YSL Club, join the Schoology group using the code posted in Schoology; First ALL meeting will be on Thursday, September 25 during prior time.
  • General note: some statements in the transcript are informal and part of class atmosphere; focus on the math content and upcoming deadlines.
  • There are two worksheets for the course: a big review for sections 2.1 and 2.2 (you won’t be able to do all of them; pick and choose; this preps for the big limits quiz) and a second worksheet given previously (day filler; more limits; not being collected as a homework assignment).
  • Online homework opened Monday; students have 120 minutes (two hours) to complete once opened; accuracy-based grading; questions can be asked but there is a time limit.
  • Friday plan: first five minutes of class, three problems from section 2.1 will be picked and written on the board; five minutes to work; that will count as your first homework quiz. If you miss Friday, you take it on the first day you return. If you do poorly, there are no retakes on the homework quiz.
  • All questions for the online homework are from the textbook.
  • The plan for the 40 limits (likely a quiz or section) is next Friday, the 26th; multiple choice questions (six to ten) may include material learned or not yet learned; the teacher will expect you to eliminate extraneous material.
  • The squeeze theorem and trig limits are the main focus in 2.1; the instructor demonstrates how to apply squeeze and provides examples.
  • The class is moving toward trig limits and a first proof today, though you won’t have to reproduce this exact proof on a test.

Squeeze Theorem: Overview and Core Idea

  • The squeeze (sandwich) theorem is used when you can bound an unknown function h(x) between two functions f(x) and g(x) whose limits are the same as x approaches a point a: if f(x) ≤ h(x) ≤ g(x) near a and lim{x→a} f(x) = lim{x→a} g(x) = L, then lim_{x→a} h(x) = L.
  • Practical use: when the middle function is hard to analyze directly, you create two simpler bounding functions whose limits match.
  • Common tactic at this level: use simple f and g such as linear bounds, or functions involving sin or cos, because these have well-known bounds.

Squeeze Theorem: Example 1 (x → 0) x · sin(1/x)

  • Goal: evaluate oxed{\, ext{lim}{x o 0} x \, rac{ ^{?}}{?}} actually the target is ext{lim}{x o 0} x \, igg| rac{}{}igg|; but the key expression is
    ext{lim}_{x o 0} x \, ext{sin}igg( rac{1}{x}igg)
  • Known bound: since -1 \le\, ext{sin}ig( frac{1}{x}ig) \, ext{(any)}\,
    abla; more usefully, for all x, -1 \, rac{}{} \, ext{sin}ig( frac{1}{x}ig) \,
    leackslashar{-1}
    le 1, so multiplying by x gives
    -|x| \, ext{le} \, x \text{sin}ig( frac{1}{x}ig) \,\le |x|.
  • Since ext{lim}{x o 0} -|x| = 0 = ext{lim}{x o 0} |x|, by the squeeze theorem,
    ext{lim}_{x o 0} x \, ext{sin}igg( rac{1}{x}igg) = 0.
  • In the transcript the simpler bound used is written as -x \, ext{le} \, x \, ext{sin}ig( frac{1}{x}ig) \,\le x (valid for x>0; with absolute value version for all x).
  • Interpretation: product of x going to 0 with a bounded oscillatory sine term tends to 0.

Squeeze Theorem: Example 2 (x → 1) (x − 1) · sin(π/(x−1))

  • Let h = x − 1; the expression becomes h \, ext{sin}igg( rac{\,oldsymbol{π}\,}{h}igg).
  • Since for all t, -1 \, o\,-1 \, ext{≤} \, ext{sin}(t) \, ext{≤} \, 1, we have for all h ≠ 0,
    - |h| \, ext{≤} \, h \, ext{sin}igg( rac{π}{h}igg) \, ext{≤} \, |h|.
  • As h → 0, both −|h| and |h| → 0, so by squeeze,
    ext{lim}_{x o 1} (x-1) \, ext{sin}igg( rac{π}{x-1}igg) = 0.
  • Practical takeaway: when you encounter a nasty oscillatory trig factor, multiply by a vanishing term to trap it with simple bounds.

Squeeze Theorem: Example 3 (f, g, h) between two known limits

  • Scenario: you’re given three functions f, g, h with h sandwiched: f(x) ≤ h(x) ≤ g(x) near a, and you know lim f = lim g = L. Then lim h = L.
  • Transcript example setup: evaluate at x = 3 for f and g to determine their values; left bound yields −16, right bound yields −20 (examples shown in the session). The corresponding third bound yields a number (8/10 or 4/5 in the example) such that the left and right give the same limit, allowing you to conclude lim h = that common value. The key idea is that you don’t need a full expression for h; only its bounding behavior and the equal limits of the outer functions.
  • Practical takeaway: use when the function is hard to compute directly and you can bound it with two simpler expressions whose limits match.

Squeeze Theorem: When to use in AP and problem types

  • It’s a “last resort” technique; not frequently used as a primary method.
  • Commonly appears when the middle function contains a sine or cosine and a vanishing multiplier, or in rare multiple-choice questions where you reason with bounds rather than explicit algebraic manipulation.
  • Instructor notes: you could apply squeeze to any of the worksheets, but not every problem is designed for squeeze; many are better tackled with factoring, conjugation, or trig identities.

Trigonometric Limits: Core Facts and Geometric Derivation

  • Foundational bounds for sine and cosine:
    • For all θ, -1 \,\le \, \sin θ \,\le \, 1 and -1 \,\le \, \cos θ \,\le \, 1.
  • Basic limits as θ → 0:
    • \lim_{θ \to 0} \frac{\sin θ}{θ} = 1.
    • \lim_{θ \to 0} \cos θ = 1.
  • The geometry-based proof idea (circle of radius 1): compare areas of a sector, an inscribed triangle, and a tangent-based triangle to bound sin θ and tan θ by θ. This leads to the classic inequality
    • for small θ > 0: \sin θ \le θ \le \tan θ.
  • From these inequalities, you derive the limit of sin θ / θ by dividing by θ and/or sin θ and using reciprocals, yielding:
    • 1 \le \frac{θ}{\sin θ} \le \frac{1}{\cos θ} \quad \Rightarrow \quad \lim_{θ \to 0} \frac{θ}{\sin θ} = 1,
    • hence \lim_{θ\to0} \frac{\sin θ}{θ} = 1.
  • A secondary result from these bounds is that the cosine limit at 0 is 1: \lim_{θ\to 0} \cos θ = 1.
  • The case of (1 − cos θ)/θ and related forms: using the same squeeze approach, you eventually deduce that
    • \lim_{θ\to 0} \frac{1 - \cos θ}{θ} = 0.
    • A common related limit is \lim_{θ\to0} \frac{1 - \cos θ}{θ^2} = \tfrac{1}{2}.
  • Practical caveat: the sine-based bounds work cleanly for expressions with sine or cosine; attempting the same reciprocal technique with cosine alone does not generally yield a simple bound, so the sine-based shortcut is a key tool.

Trig Limits: A Key Derivative Result (the limit that proves the derivative of sin)

  • Goal: prove the standard derivative identity at a point x: \lim_{h\to 0} \frac{\sin(x+h) - \sin x}{h} = \cos x.
  • Start from the sine addition formula:
    • \sin(x+h) = \sin x \cos h + \cos x \sin h.
  • Subtract sin x and rewrite:
    • \sin(x+h) - \sin x = \sin x(\cos h - 1) + \cos x \sin h.
  • Divide by h:
    • \frac{\sin(x+h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}.
  • Limits as h → 0:
    • \lim_{h\to0} \frac{\sin h}{h} = 1.
    • \lim_{h\to0} \frac{\cos h - 1}{h} = 0 (via standard small-angle approximations or using the fact that cos h − 1 ≈ −h^2/2 for small h).
  • Therefore, the limit is \cos x.
  • Takeaways: this derivative limit is a central result you’ll memorize and rely on in calculus; it connects trig identities with limits and underpins differentiation of sine.

Trig Limit Shortcuts and Practical Rules

  • Key shortcut: the reciprocal trick for sine-limits is powerful:
    • If you know \lim{θ\to 0} \frac{\sin θ}{θ} = 1, then \lim{θ\to 0} \frac{θ}{\sin θ} = 1 as well.
    • This yields a handy bound: \cos θ \,\le\, \frac{\sin θ}{θ} \,\le\, 1. Taking limits gives the same result for the sine limit and shows cos θ → 1.
  • Important caveat: the reciprocal trick works cleanly for sine expressions of the form sin(kθ)/(kθ) or variants with the same multiplier in numerator and denominator. It does not directly apply to cosine in the same way.
  • Strategy for trig limits: look for the two established shortcuts (reciprocal sine bounds and cosine bounds) and then check if a direct algebraic manipulation (using addition formulas) can reduce the expression to one of these standard limits.

Worked Examples: Practical Trig Limits

  • Example A: lim_{h→0} [sin(4h)]/h
    • Rewrite as: \frac{\sin(4h)}{h} = 4 \cdot \frac{\sin(4h)}{4h}.
    • Use the standard limit: \lim_{u\to0} \frac{\sin u}{u} = 1, with u = 4h, so the limit is 4 \cdot 1 = 4.
    • Important note: you cannot simply pull the 4 inside the sine; you must treat it as a constant multiplier outside and use the transformation shown.
  • Example B: lim_{x→0} [sin(5x) cos(5x)]/(6x)
    • Rewrite as: \frac{\sin(5x) \cos(5x)}{6x} = \left( \frac{\sin(5x)}{5x} \right) \left( \cos(5x) \right) \left( \frac{5}{6} \right).
    • As x→0: \frac{\sin(5x)}{5x} \to 1 and \cos(5x) \to 1. Thus the limit is 1 \cdot 1 \cdot \frac{5}{6} = \frac{5}{6}.
    • Alternative form: you can also write as \frac{\sin(5x)}{5x} \cdot \frac{\cos(5x)}{1} \cdot \frac{5}{6} and apply the same limits.
  • Example C (using the derivative result): lim_{x→a} [\sin(x+h) - \sin x]/h = cos(a)
    • This is the general form of the derivative of sin at x = a; the same steps as in the derivative proof apply.

Summary of Core Takeaways from 2.1 and 2.2 Content

  • Squeeze theorem: construct upper and lower bounding functions that trap the target function; ensure the two bounds have the same limit.
  • When dealing with expressions that include sin or cos multiplied by a vanishing factor, consider bounding the non-sine factor and use limits of sine/cosine to pin down the limit.
  • For trig limits, memorize the key limits:
    • \lim_{θ\to0} \frac{\sin θ}{θ} = 1,
    • \lim_{θ\to0} \cos θ = 1,
    • \lim_{θ\to0} \frac{1 - \cos θ}{θ} = 0,
    • and the related \lim_{θ\to0} \frac{1 - \cos θ}{θ^2} = \tfrac{1}{2}.
  • Common trick: whenever you see sin(kx)/(kx) or cos-related forms, look for a way to factor, convert to a standard limit, and apply the squeeze or limit rules cleanly.
  • Always verify whether a trig limit can be handled with a simple algebraic manipulation, a trig identity (like the sine addition formula), or a geometric bound; avoid forcing a method that doesn’t fit.

Final Notes on Practice and Homework Strategy

  • The upcoming quizzes will emphasize the 2.1 material (limits and trig limits) and a short 2.2 section; on Fridays, the instructor will select a few problems from 2.1 to solve on the board and students will have five minutes to work, as a preparation for the first homework quiz.
  • If you miss the Friday session, you will take the quiz on the next class session you attend; there are no retakes for the homework quiz.
  • Use the textbook for all online homework and rely on the standard limit results and trig identities to work through the problems; calculators and tables can help, but the goal is to justify the steps using fundamental limits and inequalities.
  • The instructor highlighted the importance of not overrelying on graphing calculator intuition for limit problems, especially around points where the function is not defined or involves an indeterminate form like 0/0.

Quick Reference Formulas (LaTeX)

  • Bounds and squeeze for x → 0 with sin(1/x):
    -\,|x| \le x \sin\left(\frac{1}{x}\right) \le |x|,
    \lim_{x\to0} x \sin\left(\frac{1}{x}\right) = 0.
  • Example 2 bound:
    -\,|h| \le h \sin\left(\frac{\pi}{h}\right) \le |h|,
    \lim_{h\to0} h \sin\left(\frac{\pi}{h}\right) = 0.
  • Sine limit via squeeze:
    \lim{\theta\to 0} \frac{\sin \theta}{\theta} = 1. Then \lim{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \Rightarrow \lim_{\theta\to 0} \frac{\theta}{\sin\theta} = 1.
  • Derivative of sin via limit:
    \lim_{h\to 0} \frac{\sin(x+h) - \sin x}{h} = \cos x.
    Expansion: \sin(x+h) - \sin x = \sin x(\cos h - 1) + \cos x \sin h.
  • Example limits:
    \lim{h\to 0} \frac{\sin(4h)}{h} = 4, \lim{x\to 0} \frac{\sin(5x) \cos(5x)}{6x} = \frac{5}{6}.