Calculus Study Notes: Limit Laws, Algebraic Limit Evaluation, and the Squeeze Theorem

Introductory Dialogue and Informal Exchange

• Personal conversation regarding severe physical symptoms (leg cramps described as feeling like a "vice grip").
• Discussion of concurrent menstrual cycles among peers, specifically mentioning Bristintha, McKenna, and Katie.

Fundamental Concept of Limits

• Definition: A limit evaluates the behavior of the outputs of a function $f(x)$ as the inputs $x$ get arbitrarily close to a target value $a$. The primary objective is determining whether the outputs approach a single real number value.
• Analytical Approaches:
  • Graphical analysis.
  • Numerical tables.
  • Algebraic techniques (the mathematically precise method).
• Basic Limits:
  • Constant Function Limit:     $\lim_{x \to a} k = k$
  • Where $k$ is a constant real number. As the input $x$ approaches $a$, the output remains constantly k$.\n * Identity Function Limit:\n    \lim_{x \to a} x = a\n * Represents the diagonal identity line y = x$. As inputs approach$a$, the outputs directly approach$a$.
  • Both foundational limit rules can be formally proven using the precise $(\epsilon, \delta)$ definition of a limit, but are accepted as fundamental properties.

The Limit Laws

• Assumptions: Assume that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist and both equal finite real numbers.
• Sum and Difference Law:   $\lim_{x \to a} (f(x) \pm g(x)) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$
  • Allows breaking down complex arithmetic expressions into simpler, individual limits.
• Constant Multiple Law:   $\lim_{x \to a} (c \cdot f(x)) = c \cdot \lim_{x \to a} f(x)$
  • Constant factors can be factored outside of the limit operator.
• Product Law:   $\lim_{x \to a} (f(x) \cdot g(x)) = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right)$
• Quotient Law:   $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$
  • Restriction: $\lim_{x \to a} g(x) \neq 0$ because division by zero is mathematically undefined.
• Root Law:   $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}$

Application of Limit Laws and Direct Substitution

• Step-by-step Law Expansion Example: Evaluate $\lim_{x \to 4} (x^2 - 5x + 3)$.
  • Step 1 (Sum and Difference Law):     $\lim_{x \to 4} (x^2 - 5x + 3) = \lim_{x \to 4} x^2 - \lim_{x \to 4} 5x + \lim_{x \to 4} 3$
  • Step 2 (Power, Constant Multiple, and Constant Laws):     $\lim_{x \to 4} x^2 = \left(\lim_{x \to 4} x\right)^2$$\lim_{x \to 4} 5x = 5 \cdot \lim_{x \to 4} x$$\lim_{x \to 4} 3 = 3$
  • Step 3 (Evaluating Identity Limits):     $\left(\lim_{x \to 4} x\right)^2 = (4)^2 = 16$$5 \cdot \lim_{x \to 4} x = 5 \cdot 4 = 20$
  • Step 4 (Arithmetic Computation):     $16 - 20 + 3 = -1$
• Direct Substitution Property:
  • Polynomial Rule: For any polynomial function $P(x)$,     $\lim_{x \to a} P(x) = P(a)$
  • Rational Function Rule: For any rational function $R(x) = \frac{P(x)}{Q(x)}$,     $\lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}$     provided that $Q(a) \neq 0$ and the value evaluates to a real number.
  • Necessity of Notation: While direct substitution yields results directly for continuous polynomial and rational functions, formal limit techniques are essential when substitution results in undefined or indeterminate forms.

Algebraic Techniques: Factoring and Indeterminate Forms

• The Indeterminate Form $\frac{0}{0}$:
  • An expression of the form $\frac{0}{0}$ is not a number, does not equal zero, and is completely undefined. It signifies that the limit cannot be determined without prior algebraic simplification.
• The Factor Theorem Connection:
  • According to the Factor Theorem from pre-calculus, if evaluating a polynomial at $x = r$ yields zero, then $(x - r)$ is guaranteed to be a linear factor of that polynomial.
• Example 1 (Quadratic Denominator): Evaluate $\lim_{x \to -5} \frac{x + 5}{x^2 + 3x - 10}$.
  • Direct Substitution Attempt:     $\frac{-5 + 5}{(-5)^2 + 3(-5) - 10} = \frac{0}{25 - 15 - 10} = \frac{0}{0}$
  • Factoring Denominator: Since plugging in $-5$ gives zero, $(x - (-5)) = (x + 5)$ must be a factor.     $x^2 + 3x - 10 = (x + 5)(x - 2)$
  • Canceling Common Factor:     $\lim_{x \to -5} \frac{x + 5}{(x + 5)(x - 2)} = \lim_{x \to -5} \frac{1}{x - 2}$
  • Direct Substitution into Simplified Expression:     $\frac{1}{-5 - 2} = -\frac{1}{7}$
• Example 2 (Cubic Numerator): Evaluate $\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$.
  • Direct Substitution Attempt:     $\frac{1^3 - 1}{1^2 - 1} = \frac{0}{0}$
  • Factoring Numerator and Denominator:     $x^2 - 1 = (x - 1)(x + 1)$$x^3 - 1 = (x - 1)(x^2 + x + 1)$
  • Canceling Common Factor $(x - 1)$:     $\lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x^2 + x + 1}{x + 1}$
  • Direct Substitution into Simplified Expression:     $\frac{1^2 + 1 + 1}{1 + 1} = \frac{3}{2}$

Algebraic Techniques: Multiplying by the Conjugate

• Application: Used when expressions contain radical terms (square roots) that evaluate to the indeterminate form $\frac{0}{0}$.
• Conjugate Definition: For an expression $a + \sqrt{b}$, its conjugate is $a - \sqrt{b}$. Multiplying conjugate pairs produces a difference of squares structure, eliminating radical terms.
• Example: Evaluate $\lim_{x \to -3} \frac{2 - \sqrt{x^2 - 5}}{x + 3}$.
  • Direct Substitution Attempt:     $\frac{2 - \sqrt{(-3)^2 - 5}}{-3 + 3} = \frac{2 - \sqrt{9 - 5}}{0} = \frac{2 - 2}{0} = \frac{0}{0}$
  • Conjugate Multiplication: Multiply numerator and denominator by $(2 + \sqrt{x^2 - 5})$.     $\lim_{x \to -3} \frac{\left(2 - \sqrt{x^2 - 5}\right)\left(2 + \sqrt{x^2 - 5}\right)}{(x + 3)\left(2 + \sqrt{x^2 - 5}\right)}$
  • Simplifying the Numerator:     $(2)^2 - \left(\sqrt{x^2 - 5}\right)^2 = 4 - (x^2 - 5) = 4 - x^2 + 5 = 9 - x^2$
  • Factoring Simplified Numerator:     $9 - x^2 = (3 - x)(3 + x)$
  • Canceling Common Factor: Note that $(3 + x)$ and $(x + 3)$ are equivalent.     $\lim_{x \to -3} \frac{(3 - x)(3 + x)}{(x + 3)\left(2 + \sqrt{x^2 - 5}\right)} = \lim_{x \to -3} \frac{3 - x}{2 + \sqrt{x^2 - 5}}$
  • Direct Substitution into Simplified Expression:     $\frac{3 - (-3)}{2 + \sqrt{(-3)^2 - 5}} = \frac{6}{2 + \sqrt{9 - 5}} = \frac{6}{2 + 2} = \frac{6}{4} = \frac{3}{2}$

Algebraic Techniques: Clearing Complex Fractions

• Definition: A complex fraction is an expression containing fractions nested within a larger fraction.
• Method: Multiply the primary numerator and denominator by the common denominator of the inner fractions to eliminate nested denominators.
• Example: Evaluate $\lim_{x \to 1} \frac{\frac{1}{x} - 1}{x - 1}$.
  • Direct Substitution Attempt:     $\frac{\frac{1}{1} - 1}{1 - 1} = \frac{0}{0}$
  • Clearing Nested Fraction: Multiply top and bottom by $x$.     $\lim_{x \to 1} \frac{\left(\frac{1}{x} - 1\right) \cdot x}{(x - 1) \cdot x} = \lim_{x \to 1} \frac{1 - x}{x(x - 1)}$
  • Factoring Negations: Recognize that $(1 - x) = -1(x - 1)$.     $\lim_{x \to 1} \frac{-1(x - 1)}{x(x - 1)}$
  • Canceling Common Factor:     $\lim_{x \to 1} \frac{-1}{x}$
  • Direct Substitution into Simplified Expression:     $\frac{-1}{1} = -1$

Precision in Mathematical Notation

• Equality Chains: Limits must be written as continuous chains of equal statements.
• Retention of Limit Symbol: The $\lim_{x \to a}$ operator must be explicitly re-written at every step until direct numerical substitution is performed.
• Termination of Limit Symbol: Once numbers are substituted for all variables, the limit notation must be dropped.

The Squeeze Theorem

• Theorem Statement: Suppose $f(x)$, $g(x)$, and $h(x)$ satisfy:   $f(x) \le g(x) \le h(x)$   for all $x$ near $a$ (except possibly at $a$). If:   $\lim_{x \to a} f(x) = L$$\lim_{x \to a} h(x) = L$   then:   $\lim_{x \to a} g(x) = L$
• Intuitive Interpretation: The middle function $g(x)$ is squeezed between lower bound $f(x)$ and upper bound $h(x)$. When both outer bounds converge to the same value $L$, $g(x)$ is forced to converge to $L$.

Applications of the Squeeze Theorem

• Problem 1: Evaluate $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

  • Analysis: Direct substitution fails because $\frac{1}{0}$ is undefined, causing $\sin\left(\frac{1}{x}\right)$ to oscillate infinitely without approaching a limit.
  • Bounding the Oscillating Function:     $-1 \le \sin\left(\frac{1}{x}\right) \le 1$
  • Multiplying by Non-negative Term $x^2$:     $-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$
  • Computing Limits of Outer Functions:     $\lim_{x \to 0} (-x^2) = 0$$\lim_{x \to 0} x^2 = 0$
  • Conclusion: By the Squeeze Theorem,     $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

• Problem 2: Evaluate $\lim_{x \to 0} x \cos\left(\frac{1}{x}\right)$.

  • Bounding Trigonometric Function:     $-1 \le \cos\left(\frac{1}{x}\right) \le 1$
  • Boundary Selection for Negative Input Values: Multiplying directly by $x$ fails for negative values ($x < 0$) because multiplying an inequality by a negative number reverses the inequality direction.
  • Absolute Value Bounding:     $-|x| \le x \cos\left(\frac{1}{x}\right) \le |x|$
  • Computing Limits of Outer Functions:     $\lim_{x \to 0} (-|x|) = 0$$\lim_{x \to 0} |x| = 0$
  • Conclusion: By the Squeeze Theorem,     $\lim_{x \to 0} x \cos\left(\frac{1}{x}\right) = 0$

Questions & Discussion

• Procedure regarding step-by-step limit law application:
  • Question: Must every limit evaluation explicitly break down terms using individual limit laws?
  • Answer: Unless a problem explicitly states "use the limit laws", steps may be combined, and direct substitution or algebraic properties may be applied immediately.
• Casual Exclamation:
  • Comment: "What the fuck is a kilometer?"
  • Context: Remark made at the conclusion of class discussion referencing upcoming topics on inclinement/continuity.