Comprehensive Study Notes on Limits, Vertical Asymptotes, and the Squeeze Theorem

Rational Function Limits and Vertical Asymptotes

• Initial Limit Evaluation and Direct Substitution Failure:

  • When evaluating the limit:         $\lim_{h \to 0} \left( \frac{3}{h} - \frac{4}{h+2} \right)$
  • Direct substitution of $h = 0$ into the expression yields $\frac{3}{0} - \frac{4}{2}$.
  • Since $\frac{3}{0}$ is undefined, direct substitution fails. Treating an undefined expression minus a defined value as a standard algebraic calculation is mathematically invalid.

• Finding the Least Common Denominator (LCD):

  • To evaluate limits involving subtracted or added rational expressions that are individually undefined, combine the terms into a single rational expression using a Least Common Denominator (LCD).
  • Rule of Thumb for LCDs:
    • If two denominators share no common factors or multiples, the LCD is simply their product.
    • Arithmetic Analogy 1: For $\frac{2}{3} - \frac{7}{8}$, the denominators $3$ and $8$ are not multiples of each other. The LCD is $3 \times 8 = 24$.
    • Arithmetic Analogy 2: For $\frac{2}{3} - \frac{1}{9}$, the denominators $3$ and $9$ are multiples. The LCD is $9$. Using $3 \times 9 = 27$ yields a valid common denominator, but it is not the least common denominator and results in unnecessarily large numbers.
  • For the denominators $h$ and $h+2$, neither is a multiple of the other, so the LCD is $h(h+2)$.

• Algebraic Manipulation Step-by-Step:

  • Multiply the first fraction by $\frac{h+2}{h+2}$:         $\frac{3(h+2)}{h(h+2)}$
  • Multiply the second fraction by $\frac{h}{h}$:         $\frac{4h}{h(h+2)}$
  • Combine the numerators over the common denominator:         $\lim_{h \to 0} \frac{3(h+2) - 4h}{h(h+2)}$
  • Distribute the $3$ in the numerator:         $\lim_{h \to 0} \frac{3h + 6 - 4h}{h(h+2)}$
  • Combine like terms ($3h - 4h = -h$):         $\lim_{h \to 0} \frac{-h + 6}{h(h+2)}$

• Re-evaluating the Limit:

  • Substitute $h = 0$ into the simplified expression:
    • Numerator: $-0 + 6 = 6$
    • Denominator: $0(0 + 2) = 0$
  • The substitution yields $\frac{6}{0}$.

Classification of Limit Results at a Point

• When substituting a target value $a$ directly into a function during a limit evaluation, four distinct structural outcomes can occur:
  1. $\frac{\text{nonzero number}}{\text{nonzero number}}$
     • The expression simplifies directly to a single real number; the limit is fully evaluated.
  2. $\frac{0}{\text{nonzero number}}$
     • The limit equals $0$; no further algebraic work is required.
  3. $\frac{0}{0}$
     • Represents an indeterminate form, which geometrically corresponds to a removable discontinuity (hole).
     • Requires further algebraic manipulation (e.g., factoring, finding common denominators, or multiplying by conjugates) to cancel common factors before re-evaluating.
  4. $\frac{\text{nonzero number}}{0}$
     • Represents an undefined form that geometrically corresponds to a vertical asymptote.
     • The two-sided limit does not exist as a real finite number. Further analysis using graphs or tables is required to determine whether one-sided limits approach $+\infty$ or $-\infty$.

Graphical and Tabular Analysis of Vertical Asymptotes

• Evaluating Asymptotic Behavior:

  • Because direct substitution into $\lim_{h \to 0} \frac{-h + 6}{h(h+2)}$ yields $\frac{6}{0}$, a vertical asymptote exists at $h = 0$.
  • Note: A second vertical asymptote exists at $h = -2$, but it is irrelevant to the behavior near $h = 0$.

• Calculator Input and Efficiency:

  • To analyze the graph or table, enter either the original unsimplified function or the simplified version $\frac{-x + 6}{x^2 + 2x}$ into the calculator, using the standard variable key $x$.
  • Calculator Tip: Use the 2nd [INS] (Insert) function located above the DEL key to insert missing parentheses without needing to retype an entire line.

• One-Sided Limit Evaluation:

  • Right-Hand Limit ($h \to 0^+$): As $h$ approaches $0$ from positive values, the graph trends upward toward positive infinity:         $\lim_{h \to 0^+} f(h) = \infty$
  • Left-Hand Limit ($h \to 0^-$): As $h$ approaches $0$ from negative values, the graph trends downward toward negative infinity:         $\lim_{h \to 0^-} f(h) = -\infty$

• Table Verification vs. Graphical Pitfalls:

  • Relying exclusively on a graphing window can lead to visual misinterpretations if the window scale is restricted. For example, setting $Y_{\text{min}} = -5$ cuts off lower graph behavior, making it incorrectly appear as if both sides go toward positive infinity.
  • Numerical evaluation via a calculator table provides clear, unambiguous directional trends:
    • Positive Inputs ($h \to 0^+$):
      • $h = 0.1 \implies f(h) = 28$
      • $h = 0.01 \implies f(h) = 298$
      • $h = 0.001 \implies f(h) = 2998$
      • Values grow infinitely positive ($\to \infty$).
    • Negative Inputs ($h \to 0^-$):
      • $h = -0.1 \implies f(h) = -32$
      • $h = -0.01 \implies f(h) = -302$
      • $h = -0.001 \implies f(h) = -3002$
      • Values grow infinitely negative ($\to -\infty$).

• Final Two-Sided Limit Conclusion:

  • Since the right-hand limit approaches $\infty$ and the left-hand limit approaches $-\infty$, the two-sided limit does not exist:         $\lim_{h \to 0} \left( \frac{3}{h} - \frac{4}{h+2} \right) = \text{DNE}$

The Squeeze Theorem (Sandwich Theorem)

• Core Concept and Purpose:

  • The Squeeze Theorem (also known as the Sandwich Theorem) is an algebraic technique used to compute limits of complex functions that cannot be simplified using standard algebraic methods.
  • Conceptual Metaphor: If an interior function is bounded between two boundary functions, and both boundary functions approach the exact same limit value $L$ as $x \to a$, the interior function is squeezed/sandwiched into approaching $L$ as well.

• Formal Mathematical Definition:

  • Let $f(x)$, $g(x)$, and $h(x)$ be functions such that:         $g(x) \le f(x) \le h(x)$         for all $x$ in an open interval around a point $a$ (except possibly at $a$ itself).
  • If the limits of the outer boundary functions $g(x)$ and $h(x)$ at $x = a$ exist and equal $L$:         $\lim_{x \to a} g(x) = L$$\lim_{x \to a} h(x) = L$
  • Then the limit of the inner function $f(x)$ as $x \to a$ must also equal $L$:         $\lim_{x \to a} f(x) = L$

• Logical Representation:

  • Substituting the limit value into the bound inequality yields:         $L \le \lim_{x \to a} f(x) \le L$
  • Because $\lim_{x \to a} f(x)$ is simultaneously greater than or equal to $L$ and less than or equal to $L$, it must equal $L$.

Application of Squeeze Theorem with Given Boundary Functions

• Problem Statement:

  • Suppose $-x^2 + 6x - 13 \le f(x) \le \cos(\pi x) - 3$. Find $\lim_{x \to 3} f(x)$.

• Boundary Function Identification:

  • Lower boundary function: $g(x) = -x^2 + 6x - 13$
  • Upper boundary function: $h(x) = \cos(\pi x) - 3$

• Step-by-Step Boundary Limit Calculations:

  1. Evaluate Lower Bound Limit as $x \to 3$:         $\lim_{x \to 3} (-x^2 + 6x - 13) = -(3)^2 + 6(3) - 13$$\lim_{x \to 3} (-x^2 + 6x - 13) = -9 + 18 - 13 = -4$
  2. Evaluate Upper Bound Limit as $x \to 3$:         $\lim_{x \to 3} (\cos(\pi x) - 3) = \cos(3\pi) - 3$
     • Unit Circle Recall: $\cos(0) = 1$, $\cos(\pi) = -1$, $\cos(2\pi) = 1$, and $\cos(3\pi) = -1$.         $\lim_{x \to 3} (\cos(\pi x) - 3) = -1 - 3 = -4$

• Conclusion:

  • Since both boundary limits equal $-4$:         $-4 \le \lim_{x \to 3} f(x) \le -4$
  • By the Squeeze Theorem:         $\lim_{x \to 3} f(x) = -4$

Constructing Boundary Functions for the Squeeze Theorem

• Problem Statement:

  • Use the Squeeze Theorem to evaluate:         $\lim_{x \to 0} \left( x^2 \sin(x) + 3 \right)$

• Process for Constructing Inequality Bounds:

  • When explicit boundary functions are not provided, construct them by starting with a known bounded component of the expression.
  • The trigonometric sine function $\sin(x)$ is bounded between $-1$ and $1$ for all real values of $x$:         $-1 \le \sin(x) \le 1$

• Building the Target Expression Step-by-Step:

  1. Start with the basic bounded sine inequality:         $-1 \le \sin(x) \le 1$
  2. Multiply all three parts of the inequality by $x^2$ (since $x^2 \ge 0$, inequality signs remain unchanged):         $-x^2 \le x^2 \sin(x) \le x^2$
  3. Add $3$ to all three parts of the inequality:         $-x^2 + 3 \le x^2 \sin(x) + 3 \le x^2 + 3$

• Identified Boundary Functions:

  • Lower boundary function: $g(x) = -x^2 + 3$
  • Upper boundary function: $h(x) = x^2 + 3$

• Evaluating Boundary Limits as $x \to 0$:

  1. Lower bound:         $\lim_{x \to 0} (-x^2 + 3) = -0^2 + 3 = 3$
  2. Upper bound:         $\lim_{x \to 0} (x^2 + 3) = 0^2 + 3 = 3$

• Conclusion:

  • Since both boundary limits equal $3$:         $3 \le \lim_{x \to 0} \left( x^2 \sin(x) + 3 \right) \le 3$
  • By the Squeeze Theorem:         $\lim_{x \to 0} \left( x^2 \sin(x) + 3 \right) = 3$

Bounded vs. Unbounded Functions in Squeeze Theorem Construction

• Suitable Base Functions:

  • $\sin(x)$: bounded strictly within $[-1, 1]$.
  • $\cos(x)$: bounded strictly within $[-1, 1]$.

• Unsuitable Base Functions:

  • $\csc(x)$ and $\sec(x)$: contain vertical asymptotes and range across $(-\infty, -1] \cup [1, \infty)$, making them globally unbounded.
  • $\tan(x)$: contains periodic vertical asymptotes and ranges from $-\infty$ to $+\infty$.
  • Parabolic functions ($x^2$): bounded below by $0$, but unbounded above (approaching $\infty$).
  • Square root functions ($\sqrt{x}$): bounded below by $0$, but unbounded above (approaching $\infty$).

Questions and Clarifications

• Question: Why does $4h$ appear in the combined numerator of $\frac{3}{h} - \frac{4}{h+2}$?

  • Response: When creating the LCD of $h(h+2)$, the second fraction $\frac{4}{h+2}$ lacks $h$. Multiplying both its numerator and denominator by $h$ gives $\frac{4h}{h(h+2)}$.

• Question: Why start building the Squeeze Theorem inequality with $\sin(x)$?

  • Response: $\sin(x)$ is present in the expression and is a recognized, globally bounded function ($-1 \le \sin(x) \le 1$). Terms like $x^2$ or $+3$ are either constants or unbounded algebraic expressions.

• Question: Can functions like $\csc(x)$ or $\tan(x)$ be used to set up Squeeze Theorem bounds?

  • Response: No, because their graphs extend to $\pm\infty$ and contain vertical asymptotes, making it impossible to trap a function between two finite numerical bounds.