2.2 Algebraic Evaluation of Limits & Indeterminate Forms

2.2 Algebraic Evaluation of Limits & Indeterminate Forms

Core CLEP Concept: When direct substitution of $x = c$ into a quotient produces the indeterminate form $\frac{0}{0}$, the limit cannot be determined without algebraic intervention. The form $\frac{0}{0}$ does not mean the limit is $0$, $1$, or undefined—it means "do more algebra." On Section 1 of the CLEP exam (no calculator), four algebraic strategies are tested repeatedly.

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1. Strategy 1: Factoring and Polynomial Cancellation

By the Factor Theorem, if substituting $x = c$ into a polynomial $P(x)$ yields $P(c) = 0$, then $(x - c)$ is guaranteed to be a factor of $P(x)$. When both the numerator and denominator evaluate to $0$, the factor $(x - c)$ is present in both and must be algebraically canceled.

Essential Factoring Formulas

• Difference of Squares: $a^2 - b^2 = (a - b)(a + b)$
• Difference of Cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
• Sum of Cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
• Quadratic Grouping: $ax^2 + bx + c = ax^2 + px + qx + c$ where $p+q = b$ and $pq = ac$

Step-by-Step Worked Example: Difference of Cubes

Problem: Evaluate $\lim_{x \to 3} \frac{x^3 - 27}{2x^2 - 5x - 3}$.

Step 1: Test direct substitution.

$$\frac{3^3 - 27}{2(3)^2 - 5(3) - 3} = \frac{27 - 27}{18 - 15 - 3} = \frac{0}{0} \quad \text{(Indeterminate Form)}$$

Step 2: Factor numerator and denominator completely.

• Numerator (Difference of Cubes with $a = x, b = 3$): $$x^3 - 3^3 = (x - 3)(x^2 + 3x + 9)$$
• Denominator (Quadratic Trinomial): $$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$

Step 3: Cancel the common vanishing factor $(x - 3)$. Since the limit considers $x \neq 3$, the factor $(x - 3) \neq 0$, so division is valid:

$$\lim_{x \to 3} \frac{(x - 3)(x^2 + 3x + 9)}{(2x + 1)(x - 3)} = \lim_{x \to 3} \frac{x^2 + 3x + 9}{2x + 1}$$

Step 4: Evaluate the simplified expression by direct substitution.

$$\frac{3^2 + 3(3) + 9}{2(3) + 1} = \frac{9 + 9 + 9}{6 + 1} = \frac{27}{7}$$

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2. Strategy 2: Rationalization with Radical Conjugates

When square roots appear in a $\frac{0}{0}$ indeterminate form, factoring is rarely straightforward. Instead, multiply the numerator and denominator by the radical conjugate. The conjugate of $(\sqrt{A} - B)$ is $(\sqrt{A} + B)$, exploiting the identity:

$$(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2$$

Step-by-Step Worked Example: Single Radical

Problem: Evaluate $\lim_{x \to 0} \frac{\sqrt{x + 16} - 4}{x}$.

Step 1: Test direct substitution.

$$\frac{\sqrt{0 + 16} - 4}{0} = \frac{4 - 4}{0} = \frac{0}{0}$$

Step 2: Multiply numerator and denominator by the conjugate $(\sqrt{x + 16} + 4)$.

$$\lim_{x \to 0} \frac{(\sqrt{x + 16} - 4)(\sqrt{x + 16} + 4)}{x(\sqrt{x + 16} + 4)}$$

Step 3: Expand the numerator.

$$\lim_{x \to 0} \frac{(x + 16) - 16}{x(\sqrt{x + 16} + 4)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 16} + 4)}$$

Step 4: Cancel the common factor $x$ and evaluate.

$$\lim_{x \to 0} \frac{1}{\sqrt{x + 16} + 4} = \frac{1}{\sqrt{16} + 4} = \frac{1}{4 + 4} = \frac{1}{8}$$

Advanced CLEP Example: Double Radical Conjugate

Problem: Evaluate $\lim_{x \to 4} \frac{\sqrt{2x + 1} - 3}{\sqrt{x} - 2}$.

Here, both numerator and denominator contain radicals producing $0$. Multiply simultaneously by the conjugates of both:

$$\lim_{x \to 4} \frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)(\sqrt{2x+1}+3)}$$ $$= \lim_{x \to 4} \frac{((2x+1) - 9)(\sqrt{x}+2)}{(x - 4)(\sqrt{2x+1}+3)} = \lim_{x \to 4} \frac{2(x - 4)(\sqrt{x}+2)}{(x - 4)(\sqrt{2x+1}+3)}$$ $$= \lim_{x \to 4} \frac{2(\sqrt{x}+2)}{\sqrt{2x+1}+3} = \frac{2(\sqrt{4}+2)}{\sqrt{9}+3} = \frac{2(4)}{6} = \frac{8}{6} = \frac{4}{3}$$

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3. Strategy 3: Simplifying Complex (Nested) Fractions

A complex fraction contains fractions within its numerator or denominator. To resolve the $\frac{0}{0}$ form, multiply the overarching numerator and denominator by the Least Common Denominator (LCD) of all internal fractions.

Step-by-Step Worked Example: Difference Quotient for $f(x) = 1/x^2$

Problem: Evaluate $\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$.

Step 1: Identify the LCD of the inner fractions. The inner denominators are $(x+h)^2$ and $x^2$. The LCD is $x^2(x+h)^2$.

Step 2: Multiply numerator and denominator by the LCD.

$$\lim_{h \to 0} \frac{\left[ \frac{1}{(x+h)^2} - \frac{1}{x^2} \right] x^2(x+h)^2}{h \cdot x^2(x+h)^2} = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h x^2(x+h)^2}$$

Step 3: Expand the numerator algebraically.

$$x^2 - (x^2 + 2xh + h^2) = -2xh - h^2 = -h(2x + h)$$

Step 4: Substitute and cancel the vanishing factor $h$.

$$\lim_{h \to 0} \frac{-h(2x + h)}{h x^2(x+h)^2} = \lim_{h \to 0} \frac{-(2x + h)}{x^2(x+h)^2}$$

Step 5: Evaluate by direct substitution ($h = 0$).

$$\frac{-(2x + 0)}{x^2(x+0)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$

(Note: This proves that $\frac{d}{dx}[x^{-2}] = -2x^{-3}$, linking limit evaluations to derivative rules.)

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4. Strategy 4: Special Trigonometric Limits

Two fundamental trigonometric limits must be memorized for the CLEP exam. All trigonometric indeterminate evaluations trace back to these two anchors:

$$\mathbf{1.} \quad \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \quad \left(\text{and } \lim_{\theta \to 0} \frac{\theta}{\sin \theta} = 1\right)$$ $$\mathbf{2.} \quad \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0$$

(Note: Angles $\theta$ are always measured in radians.)

Proof Sketch of $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0$

Multiply by the conjugate $(1 + \cos \theta)$:

$$\lim_{\theta \to 0} \frac{(1 - \cos \theta)(1 + \cos \theta)}{\theta(1 + \cos \theta)} = \lim_{\theta \to 0} \frac{1 - \cos^2 \theta}{\theta(1 + \cos \theta)} = \lim_{\theta \to 0} \frac{\sin^2 \theta}{\theta(1 + \cos \theta)}$$ $$= \left( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \right) \cdot \left( \lim_{\theta \to 0} \frac{\sin \theta}{1 + \cos \theta} \right) = 1 \cdot \frac{0}{1 + 1} = 1 \cdot 0 = 0$$

The Coefficient Scaling Rule

For any non-zero constants $a$ and $b$:

$$\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b} \lim_{x \to 0} \frac{\sin(ax)}{ax} = \frac{a}{b} \cdot 1 = \frac{a}{b}$$

Worked Example: Trigonometric Limits with Tangent

Problem: Evaluate $\lim_{x \to 0} \frac{\tan(5x)}{\sin(3x)}$.

Rewrite $\tan(5x) = \frac{\sin(5x)}{\cos(5x)}$ and separate into fundamental limit forms:

$$\lim_{x \to 0} \left[ \frac{\sin(5x)}{5x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{5}{3} \cdot \frac{1}{\cos(5x)} \right]$$

As $x \to 0$:

$$1 \cdot 1 \cdot \frac{5}{3} \cdot \frac{1}{\cos(0)} = \frac{5}{3} \cdot 1 = \frac{5}{3}$$

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5. Absolute Value and Piecewise Limits

Expressions of the form $\frac{|x - c|}{x - c}$ always require computing separate left-hand and right-hand limits because:

$$|x - c| = \begin{cases} x - c & \text{if } x > c \\ -(x - c) & \text{if } x < c \end{cases}$$

Worked Example: Absolute Value Ratio

Problem: Evaluate $\lim_{x \to 4} \frac{|x - 4|}{x^2 - 16}$.

Factor denominator: $x^2 - 16 = (x - 4)(x + 4)$.

• Left-Hand Limit ($x \to 4^-$): For $x < 4$, $x - 4 < 0$, so $|x - 4| = -(x - 4)$: $$\lim_{x \to 4^-} \frac{-(x - 4)}{(x - 4)(x + 4)} = \lim_{x \to 4^-} \frac{-1}{x + 4} = -\frac{1}{8}$$
• Right-Hand Limit ($x \to 4^+$): For $x > 4$, $x - 4 > 0$, so $|x - 4| = (x - 4)$: $$\lim_{x \to 4^+} \frac{x - 4}{(x - 4)(x + 4)} = \lim_{x \to 4^+} \frac{1}{x + 4} = \frac{1}{8}$$
• Conclusion: Since $-\frac{1}{8} \neq \frac{1}{8}$, the two-sided limit Does Not Exist (DNE).

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6. Strategic Decision Summary