Calculus Notes: Limit Laws and Algebraic Techniques

Direct Substitution and Domain Restrictions

• Inaccuracy of Numerical/Table Approximations:

  • Selecting arbitrary $x$-values or table data points is an unreliable method for determining limits.
  • A function may appear to approach $0$ based on selected points, whereas it actually approaches a non-zero value such as $0.0001$

• Definition of Direct Substitution:

  • If $f$ is a polynomial or rational function and $a$ is in the domain of $f$, then:     $\lim_{x \to a} f(x) = f(a)$
  • A value $a$ is in the domain of $f$ if $f(a)$ is well-defined (e.g., no division by zero, no negative values under an even radical).
  • If $a$ is in the domain, the limit is evaluated simply by plugging $a$ into the function.

• Examples of Direct Substitution:

  • Example 1: Evaluate $\lim_{x \to 1} \frac{-3}{(x-2)^2}$
  • Determine domain restrictions: The denominator equals zero when $x = 2$. Thus, $x = 2$ is not in the domain.
  • Since the limit approaches $x = 1$, and $1$ is in the domain, evaluate by direct substitution:       $\lim_{x \to 1} \frac{-3}{(x-2)^2} = \frac{-3}{(1-2)^2} = \frac{-3}{(-1)^2} = \frac{-3}{1} = -3$
  • Example 2: Evaluate $\lim_{x \to 2} \frac{2x - 4}{6 - 3x^2}$
  • Substitute $x = 2$ directly:       $\frac{2(2) - 4}{6 - 3(2)^2} = \frac{4 - 4}{6 - 3(4)} = \frac{0}{6 - 12} = \frac{0}{-6} = 0$
  • When a fraction has $0$ in the numerator and a non-zero number in the denominator, the result is $0$
  • Example 3: Evaluate $\lim_{x \to 2} \frac{\sqrt{x^2 + 5}}{\ln(x)}$
  • Substitute $x = 2$ directly:       $\frac{\sqrt{2^2 + 5}}{\ln(2)} = \frac{\sqrt{4 + 5}}{\ln(2)} = \frac{\sqrt{9}}{\ln(2)} = \frac{3}{\ln(2)}$
  • Unsimplified exact expressions (such as leaving $\ln(2)$ or $\cos(\frac{\pi}{3})$ in exact exact form) are preferred over incorrect arithmetic simplification.

Indeterminate Form and Algebraic Strategies

• Indeterminate Form Definition:

  • When direct substitution yields $\frac{0}{0}$, the limit is in indeterminate form.
  • $\frac{0}{0}$ is distinct from being undefined or equal to $0$; it signifies insufficient information to determine the behavior of the function at that point.
  • Indeterminate form requires algebraic manipulation (such as factoring, rationalizing, or simplifying complex fractions) to rewrite the expression before evaluating.

• Algebraic Technique 1: Factoring Quadratics:

  • Example 1: Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
  • Direct substitution yields $\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$ (indeterminate form).
  • Factor the numerator using difference of squares:       $\lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}$
  • Cancel the common factor $(x - 2)$:       $\lim_{x \to 2} (x + 2)$
  • Substitute $x = 2$:       $2 + 2 = 4$
  • Example 2: Evaluate $\lim_{x \to 2} \frac{x^2 - 7x + 10}{x - 2}$
  • Direct substitution yields $\frac{2^2 - 7(2) + 10}{2 - 2} = \frac{4 - 14 + 10}{0} = \frac{0}{0}$ (indeterminate form).
  • Factor the numerator by finding two numbers that multiply to $10$ and add to $-7$ (which are $-5$ and $-2$):       $\lim_{x \to 2} \frac{(x - 5)(x - 2)}{x - 2}$
  • Cancel $(x - 2)$ and apply direct substitution:       $\lim_{x \to 2} (x - 5) = 2 - 5 = -3$

• Notation Requirement:

  • The limit operator ($\lim_{x \to a}$) must be explicitly written at every step of algebraic simplification.
  • The limit operator is dropped only at the step where direct numerical substitution is performed.

Limit Laws

• Formal Definition:

  • Let $c$ be a constant, and assume that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist.

• Sum and Difference Law:$\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$

  • Limits can be added or subtracted provided both functions are approaching the exact same value $a$

• Constant Multiple Law:$\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)$

  • A constant factor $c$ (such as $3$, $\pi$, or $-2.789$) can be factored outside or moved inside the limit.
  • Example: $\lim_{x \to 1} (3x^2) = 3 \cdot \lim_{x \to 1} (x^2)$

• Product Law:$\lim_{x \to a} [f(x) \cdot g(x)] = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right)$

  • The limit of a product equals the product of the individual limits.

Advanced Algebraic Techniques for Limits

• Algebraic Technique 2: Rationalizing Using Conjugates:

  • Applicable when encountering radical expressions resulting in $\frac{0}{0}$
  • Conjugate Rule: For an expression $\sqrt{a} + b$, its conjugate is $\sqrt{a} - b$. Multiplying them yields:     $(\sqrt{a} + b)(\sqrt{a} - b) = a - b^2$
  • Example: Evaluate $\lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}$
  • Direct substitution yields $\frac{\sqrt{9+0} - 3}{0} = \frac{0}{0}$ (indeterminate form).
  • Multiply numerator and denominator by the conjugate of the numerator, $\frac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3}$ (which equals $1$):       $\lim_{h \to 0} \left( \frac{\sqrt{9 + h} - 3}{h} \cdot \frac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3} \right)$
  • Expand the numerator:       $(\sqrt{9+h})^2 - (3)^2 = (9 + h) - 9 = h$
  • Rewrite expression:       $\lim_{h \to 0} \frac{h}{h(\sqrt{9 + h} + 3)}$
  • Cancel the common factor $h$:       $\lim_{h \to 0} \frac{1}{\sqrt{9 + h} + 3}$
  • Substitute $h = 0$:       $\frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

• Algebraic Technique 3: Simplifying Complex Fractions:

  • Applicable when limits contain nested rational functions.
  • Cross-Multiplication Rule:$\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$
  • Example: Evaluate $\lim_{x \to 4} \frac{\frac{1}{x-2} - \frac{1}{2}}{4 - x}$
  • Direct substitution yields $\frac{\frac{1}{2} - \frac{1}{2}}{4 - 4} = \frac{0}{0}$ (indeterminate form).
  • Cross-multiply the numerator terms:       $\frac{1}{x-2} - \frac{1}{2} = \frac{2 - (x - 2)}{2(x - 2)} = \frac{2 - x + 2}{2(x - 2)} = \frac{4 - x}{2(x - 2)}$
  • Treat denominator $(4 - x)$ as $\frac{4 - x}{1}$ and apply multiplication by the reciprocal:       $\lim_{x \to 4} \left( \frac{4 - x}{2(x - 2)} \cdot \frac{1}{4 - x} \right)$
  • Cancel $(4 - x)$:       $\lim_{x \to 4} \frac{1}{2(x - 2)}$
  • Substitute $x = 4$:       $\frac{1}{2(4 - 2)} = \frac{1}{2(2)} = \frac{1}{4}$

Categorization of Zero Outcomes in Limits

• Case 1: Indeterminate Form ($\frac{0}{0}$)

  • Example: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$
  • Evaluation: $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$
  • Action: Requires algebraic manipulation (factoring):     $\lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$
  • Never leave $\frac{0}{0}$ as the final answer.

• Case 2: Zero Numerator, Non-Zero Denominator ($\frac{0}{k}$ where $k \neq 0$)

  • Example: $\lim_{x \to 3} \frac{x - 3}{x + 3}$
  • Evaluation: $\frac{3 - 3}{3 + 3} = \frac{0}{6} = 0$
  • Result: Always equals $0$

• Case 3: Non-Zero Numerator, Zero Denominator ($\frac{k}{0}$ where $k \neq 0$)

  • Example: $\lim_{x \to 3} \frac{x + 3}{x - 3}$
  • Evaluation: $\frac{3 + 3}{3 - 3} = \frac{6}{0}$
  • Result: Undefined / Does Not Exist (DNE) / None.

Graphical Limit Evaluations and Limit Laws

• Applying Limit Laws Graphically:

  • Given graphical functions $f(x)$ and $g(x)$, limit laws allow distributing limits across function operations.

• Graphical Example 1: Evaluate $\lim_{x \to -2} [f(x) + 5g(x)]$

  • Distribute limit using Sum and Constant Multiple Laws:     $\lim_{x \to -2} f(x) + 5 \cdot \lim_{x \to -2} g(x)$
  • Read values from graph at $x = -2$:
  • $\lim_{x \to -2} f(x) = 1$
  • $\lim_{x \to -2} g(x) = -1$
  • Compute total:     $1 + 5(-1) = 1 - 5 = -4$

• Graphical Example 2: Evaluate $\lim_{x \to 1} [f(x) \cdot g(x)]$

  • Distribute limit using Product Law:     $\left(\lim_{x \to 1} f(x)\right) \cdot \left(\lim_{x \to 1} g(x)\right)$
  • Read values from graph at $x = 1$:
  • $\lim_{x \to 1} f(x) = 2$
  • For $g(x)$ at $x = 1$, the left-hand limit and right-hand limit are not equal, so $\lim_{x \to 1} g(x)$ Does Not Exist (DNE).
  • Result: If any constituent limit does not exist, the combined overall limit Does Not Exist (DNE).

Practice Problems

• Problem A: Evaluate $\lim_{x \to 3} (x^3 + 2x^2 - 5x + 7)$

  • Direct substitution:     $3^3 + 2(3^2) - 5(3) + 7 = 27 + 2(9) - 15 + 7 = 27 + 18 - 15 + 7 = 37$

• Problem B: Evaluate $\lim_{x \to 5} \frac{x^2 - 2x - 15}{x^2 - 25}$

  • Substitution test: $\frac{5^2 - 2(5) - 15}{5^2 - 25} = \frac{25 - 10 - 15}{25 - 25} = \frac{0}{0}$ (indeterminate).
  • Factor numerator: $(x - 5)(x + 3)$
  • Factor denominator: $(x - 5)(x + 5)$
  • Rewrite limit:     $\lim_{x \to 5} \frac{(x - 5)(x + 3)}{(x - 5)(x + 5)} = \lim_{x \to 5} \frac{x + 3}{x + 5}$
  • Direct substitution:     $\frac{5 + 3}{5 + 5} = \frac{8}{10} = \frac{4}{5} = 0.8$