Limits and Continuity - Detailed Notes

Introduction to Limits and Sine Near Zero

• The instructor begins with a limit on the board and mentions there are two ways to interpret these limits:
  • Interpretation 1: Treat sin(x) as a function and compare it locally to a linear function near zero.
  • Interpretation 2: Consider the rational function sin(x)/x and study its limit as x approaches zero.
• Sine function overview from the lecture:
  • Sinusoidal wave with amplitude between rac{-1}{1} and 1.
  • The speaker mentions a period near  the graph shows sin(x) with a period (the speaker suggests  ) π, but note that the actual period of sin(x) is 2π; the statement here reflects the lecturer's comment.
• Local behavior near zero:
  • When x is very close to 0 (from the left or from the right), the graph of sin(x) looks almost linear.
  • Concretely, sin(x) is approximately equal to x for small x, i.e. sin(x) ≈ x as x → 0.
• Key consequence: the ratio sin(x)/x tends to 1 near zero.
  • Expressed as a limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1.$
  • This implies sin(x) is locally linear with slope 1 around zero.
• Two interpretations highlighted:
  • Interpretation A: sin(x) is approximately the linear function y = x when x is near 0.
  • Interpretation B: the function sin(x)/x has the limit 1 as x → 0 (the ratio test).
• Graphical intuition:
  • The graph of sin(x) and the line y = x coincide for values of x near 0 (locally).
  • For x near 0 but not equal to 0, sin(x)/x is close to 1, reflecting the local linear approximation.
• The instructor uses this to motivate limits and to set up later squeeze and algebraic tricks.

The Squeeze Theorem: Review and a Classic Example

• The squeeze idea (as reviewed from the previous class):
  • If g(x) ≤ f(x) ≤ h(x) near a, and lim{x->a} g(x) = lim{x->a} h(x) = L, then lim_{x->a} f(x) = L.
• A classic squeeze example shown: f(x) = x^2 sin(1/x).
  • Since |sin(1/x)| ≤ 1, we have
    -x^2 ≤ x^2 sin(1/x) ≤ x^2 for all x ≠ 0.
  • As x → 0, both -x^2 and x^2 approach 0.
  • Therefore, by the squeeze theorem,
    $\lim_{x\to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.$
• Graphical interpretation:
  • The function x^2 sin(1/x) is bounded between the purple parabola (opening upward) and the negative parabola (opening downward) shown in the Desmos visualization.
• Practical takeaway:
  • The squeeze step is literally about “squeezing” the function between two bounding functions whose limits agree.
  • The more you practice, the more you develop an instinct for recognizing when squeeze applies.
• Notational and conceptual rules:
  • If g(x) ≤ f(x) ≤ h(x) and lim g(x) = lim h(x) = L, then lim f(x) = L.
  • The limit of a product equals the product of the limits when the individual limits exist: $\lim<em>{x\to a} [u(x) v(x)] = \left(\lim</em>{x\to a} u(x)\right) \left(\lim_{x\to a} v(x)\right).$
• Philosophical note the lecturer repeats:
  • Practice problem sets to solidify the squeeze technique; it’s a skill that becomes instinctive with repetition.

Manipulating Limits: The “Multiply by One” Trick and a Guiding Philosophy

• Core idea: to evaluate difficult limits, you manipulate the expression by multiplying by a cleverly chosen "one" that leaves the limit unchanged but reveals a standard limit you know (most often sin(x)/x or related).
• Example outline discussed:
  • If you have a limit involving a term that cancels as x → 0, you can introduce a factor that tends to 1 to expose known limits.
  • Conceptual rule: multiply and divide by the same quantity to create a known limit (e.g., sin(t)/t) inside the expression.
• The teacher emphasizes two key strategies:
  • Divide and conquer: break the limit into parts that you can handle separately using known limits.
  • Do what works for you: different students may find different decompositions easier.
• A concrete takeaway example described:
  • When you see a structure like something × sin(x)/x, recognize that lim_{x->0} sin(x)/x = 1 and factor accordingly.
• Practical guidance:
  • If you’re unsure, note it and ask during office hours or TA sessions; algebra helps calculus a lot.

A Trigonometric Identity to Tackle a Tough Limit

• The instructor introduces a trick using the half-angle identity:
  • Start with the double-angle identity: $\cos(2\theta) = 1 - 2\sin^2\theta.$
  • Let (\theta = x/2). Then
    $\cos x = 1 - 2\sin^2\left(\frac{x}{2}\right).$
• Consequences and manipulations:
  • Rearranging gives
    $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right).$
  • This is used to compute limits like $\lim_{x\to 0} \frac{1 - \cos x}{x^2}.$
• Two derivations to show the limit equals 1/2:
  • Derivation A (via sin(x/2)/(x/2) → 1):
  • Start from
    $\frac{1 - \cos x}{x^2} = \frac{2\sin^2\left(\frac{x}{2}\right)}{x^2} = \frac{1}{2}\left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2.$
  • As x → 0, (\frac{\sin(x/2)}{x/2} \to 1), hence the limit is $\frac{1}{2}.$
  • Derivation B (via the 1 − cos x trick with sin x / x):
  • Multiply numerator and denominator by (1 + \cos x):
    $\frac{1 - \cos x}{x^2} = \frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)} = \frac{1 - \cos^2 x}{x^2(1 + \cos x)} = \frac{\sin^2 x}{x^2(1 + \cos x)}.$
  • Express as
    $\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x}.$
  • Take limits: (\lim_{x\to 0} (\sin x / x)^2 = 1^2 = 1) and (\cos x \to 1), so (1/(1 + \cos x) \to 1/2).
  • Therefore the limit is also $\frac{1}{2}.$
• Summary result:
  • $\lim_{x\to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}.$
• The instructor also notes that the limit of sin t / t is 1 for any angle t → 0, a recurring theme in these derivations.

Continuity and the Epsilon-Delta Definition

• Transition to a fundamental concept:
  • Substitution principle: if the limit as x approaches a of f(x) is L, and f is continuous at a, then the limit equals f(a).
  • Continuity at a point a is defined by three conditions:
    1) f(a) is defined (a is in the domain of f).
    2) The left-hand limit and the right-hand limit exist and are equal: $\lim<em>{x\to a^-} f(x) = \lim</em>{x\to a^+} f(x) = L.$
    3) The limit equals the function value at a: $\lim_{x\to a} f(x) = f(a).$
• The epsilon-delta formal definition of continuity:
  • For every ε > 0 there exists δ > 0 such that if (0 < |x - a| < δ) then (|f(x) - f(a)| < ε).
  • In some texts this is stated with 0 < |x - a| < δ to emphasize that the limit is about approaching a, not necessarily at a.
• Geometric interpretation:
  • In the xy-plane, as x approaches a, the corresponding y-values f(x) stay arbitrarily close to f(a).
  • Intuitively, a continuous function has no jumps or breaks at a; the graph can be drawn without lifting the pen.
• A practical, human-friendly note from the lecturer:
  • The epsilon-delta language is precise but can feel abstract; the core idea is that small changes in x near a produce small changes in f(x) near f(a).
• A student-oriented reminder:
  • If you’re stuck, write down the epsilon-delta statement, work with inequalities, and try to derive a δ as a function of ε for simple functions (e.g., f(x) = x^2) to get intuition.
• Optional illustrative example (brief): continuity of f(x) = x^2 at a = 3 can be demonstrated by bounding |x^2 - 9| = |x - 3||x + 3| and choosing δ so that |x| is bounded; then select δ to ensure |x^2 - 9| < ε. This mirrors the general epsilon-delta approach described.

Quick Recap of Key Formulas and Concepts (LaTeX)

• Local linear approximation of sine near zero:
  • $\sin x \approx x \quad \text{as } x \to 0.$
• Limit of sin x over x:
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1.$
• Squeeze theorem (informal): if (g(x) \le f(x) \le h(x)) near a and (\lim{x\to a} g(x) = \lim{x\to a} h(x) = L), then
  • $\lim_{x\to a} f(x) = L.$
• Squeeze example: for f(x) = x^2 sin(1/x),
  • $-x^2 \le x^2 \sin(1/x) \le x^2,$
  • $\lim<em>{x\to 0} (-x^2) = \lim</em>{x\to 0} x^2 = 0,$
  • hence $\lim_{x\to 0} x^2 \sin(1/x) = 0.$
• Half-angle/trigonometric identity used for a tricky limit:
  • $\cos(2\theta) = 1 - 2\sin^2\theta$
  • with (\theta = x/2): $\cos x = 1 - 2\sin^2\left(\frac{x}{2}\right)$ and $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right).$
• Limit involving 1 - cos x:
  • Approach A: $\frac{1 - \cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \xrightarrow[x\to 0]{} \frac{1}{2}.$
  • Approach B: multiply by (1 + \cos x):
    $\frac{1 - \cos x}{x^2} = \frac{\sin^2 x}{x^2(1 + \cos x)} = \left(\frac{\sin x}{x}\right)^2 \frac{1}{1 + \cos x} \xrightarrow[x\to 0]{} 1 \cdot \frac{1}{2} = \frac{1}{2}.$
• Continuity (epsilon-delta):
  • For every (\varepsilon > 0) there exists (\delta > 0) such that if (|x - a| < \delta) then (|f(x) - f(a)| < \varepsilon).
• Geometric interpretation of continuity: small horizontal changes around a map to small vertical changes around the function value.

Practical Tips and Resources Mentioned

• Desmos is a free tool that the instructor recommends for visualizing limits and graphs.
• The importance of algebra in calculus is emphasized: without algebra, calculus can be very challenging (the instructor humorously notes this).
• If questions arise during or after class, students are encouraged to ask during office hours or via email/TA sessions to build understanding.