MATH-181 (Calculus I) Curriculum

Session 03 - Computing Limits with Algebraic Techniques

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Apply fundamental limit laws to distribute limit operators across composite algebraic expressions.
Evaluate common limits and assess the existence of two-sided limits for piecewise-defined functions.
Resolve indeterminate forms ( $\frac{0}{0}$ ) using factor-and-cancel strategies (with polynomial long division), radical conjugate rationalization, and trigonometric identities.

Micro-Lecture

1. Fundamental Limit Laws & Distributive Rules

Mathematical Foundation: When individual limits $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ exist as finite real numbers, the limit operator behaves linearly, distributing across addition (Sum Law), subtraction (Difference Law), scalar multiplication (Constant Multiple Law), products (Product Law), quotients (Quotient Law), and powers/roots (Power & Root Law).
Engineering Context: In aerospace and circuit engineering, overall system responses are often composite functions of independent telemetry channels (such as thrust $\times$ mass burn or branch voltage sums). Limit laws allow engineers to evaluate each sensor channel independently before combining them.

2. Piecewise Boundaries & Directional Convergence

Mathematical Foundation: For piecewise functions whose algebraic rules change across domain boundaries, the Existence Theorem for Two-Sided Limits dictates that an overall limit exists if and only if both the left-hand and right-hand paths converge to the identical real value ( $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$ ). When paths disagree, a jump discontinuity occurs.
Engineering Context: Piecewise models represent sudden switching behaviors in electronics (such as surge protection clamp circuits) or physical material fracture limits where stress drops abruptly to zero upon structural yield.

3. Indeterminate Forms ( $\frac{0}{0}$ ) & Algebraic Resolution

Mathematical Foundation: Direct substitution yielding $\frac{0}{0}$ indicates a removable discontinuity (a hole) caused by a shared factor $(x - a)$ . Algebraic techniques—factor and cancel (using traditional factoring or polynomial long division), radical conjugate rationalization, and trigonometric identities—isolate and cancel the zero-inducing factor $(x - a)$ before re-evaluating.
Engineering Context: Fluid control valves and robotic arm motion controllers frequently encounter mathematical singularities at calibration thresholds. Algebraic resolution extracts the true continuous physical rate (flow velocity or tip deflection) without division-by-zero crashes.

Skill Block 1

Worked Example A: Evaluating Combined Functions Using Limit Laws

Given the baseline conditions $\lim_{x \to 1} f(x) = 2$ , $\lim_{x \to 1} g(x) = -3$ , and $\lim_{x \to 1} h(x) = 5$ , evaluate each of the following limits: (a) $\lim_{x \to 1} [f(x) + g(x)]$
(b) $\lim_{x \to 1} [g(x) - h(x)]$
(c) $\lim_{x \to 1} [f(x) \cdot h(x)]$
(d) $\lim_{x \to 1} [3f(x)]^2$
(e) $\lim_{x \to 1} \frac{g(x)}{h(x)}$

Worked Solution:

Part (a) — Sum Law: Apply the Sum Law to distribute the limit across the sum: $\lim_{x \to 1} [f(x) + g(x)] = \lim_{x \to 1} f(x) + \lim_{x \to 1} g(x) = 2 + (-3) = -1$
Part (b) — Difference Law: Apply the Difference Law: $\lim_{x \to 1} [g(x) - h(x)] = \lim_{x \to 1} g(x) - \lim_{x \to 1} h(x) = -3 - 5 = -8$
Part (c) — Product Law: Apply the Product Law: $\lim_{x \to 1} [f(x) \cdot h(x)] = \left(\lim_{x \to 1} f(x)\right) \cdot \left(\lim_{x \to 1} h(x)\right) = 2 \cdot 5 = 10$
Part (d) — Power & Constant Multiple Laws: Pull the constant scalar out and bring the limit inside the square: $\lim_{x \to 1} [3f(x)]^2 = \left[ 3 \cdot \lim_{x \to 1} f(x) \right]^2 = [3 \cdot 2]^2 = 6^2 = 36$
Part (e) — Quotient Law: Because the denominator limit is non-zero ( $\lim_{x \to 1} h(x) = 5 \neq 0$ ), apply the Quotient Law: $\lim_{x \to 1} \frac{g(x)}{h(x)} = \frac{\lim_{x \to 1} g(x)}{\lim_{x \to 1} h(x)} = \frac{-3}{5} = -\frac{3}{5}$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, under what conditions are the fundamental limit laws (like the Sum Law and Product Law) valid? When can they not be applied directly?
(b) In Worked Example A(d), how were the Constant Multiple Law and Power Law combined to compute $\lim_{x \to 1} [3f(x)]^2$ ?
(c) Review the linked Engineering Applications section. How does an electrical engineer use the Sum and Difference Laws to analyze the total voltage of a multi-branch circuit mesh?

Solution Key:

(a) As stated in the Conceptual Framework and Key Equations, Limit Laws are valid if and only if each individual limit $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exists and evaluates to a finite real number. If a component yields $\pm\infty$ or $\frac{0}{0}$ , the expression must first be algebraically simplified.
(b) In Worked Example A, part (d) applies the Power Law to move the limit inside the square ( $\lim [u(x)]^2 = [\lim u(x)]^2$ ), and the Constant Multiple Law pulls out the factor $3$ ( $[3 \lim f(x)]^2$ ). Substituting $\lim f(x) = 2$ gives $[3(2)]^2 = 6^2 = 36$ .
(c) In the linked Engineering Applications ("Electrical Circuit Mesh Network"), the net voltage $V_{\text{net}}(t) = V_1(t) + V_2(t) - V_3(t)$ is evaluated by applying the Sum Law and Difference Law to evaluate each branch limit independently ( $12 + 5 - 3 = 14\text{ V}$ ), verifying steady-state electrical safety without inductive spikes.

Show your steps and calculations here.

Practice Problem 2

Given the baseline conditions $\lim_{x \to 2} f(x) = 4$ , $\lim_{x \to 2} g(x) = -2$ , and $\lim_{x \to 2} h(x) = 0$ , evaluate each of the following limits using the Limit Laws, or explain why a law cannot be applied:

(a) $\lim_{x \to 2} [2f(x) - 3g(x)]$
(b) $\lim_{x \to 2} \frac{f(x) \cdot g(x)}{f(x) + g(x)}$
(c) $\lim_{x \to 2} \sqrt{f(x) - g(x) + 3}$
(d) Why can't we use the Quotient Law to evaluate $\lim_{x \to 2} \frac{f(x)}{h(x)}$ ?

Solution Key:

Part (a): $\lim_{x \to 2} [2f(x) - 3g(x)] = 2\lim_{x \to 2} f(x) - 3\lim_{x \to 2} g(x) = 2(4) - 3(-2) = 8 + 6 = 14$
Part (b): First check the denominator limit: $\lim_{x \to 2} [f(x) + g(x)] = 4 + (-2) = 2 \neq 0$ . Applying the Quotient and Product Laws: $\lim_{x \to 2} \frac{f(x) \cdot g(x)}{f(x) + g(x)} = \frac{\left(\lim_{x \to 2} f(x)\right) \cdot \left(\lim_{x \to 2} g(x)\right)}{\lim_{x \to 2} f(x) + \lim_{x \to 2} g(x)} = \frac{(4)(-2)}{4 + (-2)} = \frac{-8}{2} = -4$
Part (c): Applying the Root Law and Sum/Difference Laws: $\lim_{x \to 2} \sqrt{f(x) - g(x) + 3} = \sqrt{\lim_{x \to 2} f(x) - \lim_{x \to 2} g(x) + \lim_{x \to 2} 3} = \sqrt{4 - (-2) + 3} = \sqrt{9} = 3$
Part (d): The Quotient Law states $\lim \frac{f(x)}{h(x)} = \frac{\lim f(x)}{\lim h(x)}$ if and only if $\lim h(x) \neq 0$ . Since $\lim_{x \to 2} h(x) = 0$ , dividing by zero is undefined, so the Quotient Law cannot be applied.

Show your steps and calculations here.

Practice Problem 3

Evaluate the limit:

\lim_{x \to 0} (x - 8)^{2/3}

Solution Key:

Apply the Power & Root Law: Because polynomial expressions are continuous, bring the limit operator inside the power: $\lim_{x \to 0} (x - 8)^{2/3} = \left[\lim_{x \to 0} (x - 8)\right]^{2/3} = (0 - 8)^{2/3} = (-8)^{2/3}$
Evaluate the Fractional Power: Recall that $a^{n/m} = (\sqrt[m]{a})^n$ . Take the cube root first, then square: $(-8)^{2/3} = \left(\sqrt[3]{-8}\right)^2 = (-2)^2 = 4$

Show your steps and calculations here.

Regroup 1

Emphasize that Limit Laws provide the formal algebraic justification for evaluating functions without graphs or tables.
Reiterate that Limit Laws cannot be applied if an individual piece evaluates to $\frac{0}{0}$ or $\pm\infty$ .

Skill Block 2

Worked Example A: Evaluating Limits of Piecewise Functions at Boundary Points

Consider the piecewise-defined function:

f(x) = \begin{cases} 0 & : x < -5 \\ \sqrt{25 - x^2} & : -5 \le x < 4 \\ 5x & : x \ge 4 \end{cases}

Determine the following limits: (a) $\lim_{x \to -5^+} f(x)$
(b) $\lim_{x \to -5^-} f(x)$
(c) $\lim_{x \to -5} f(x)$
(d) $\lim_{x \to 4^+} f(x)$
(e) $\lim_{x \to 4^-} f(x)$
(f) $\lim_{x \to 4} f(x)$

Worked Solution:

Analyze Boundary $x = -5$ :
- (a) Right-sided limit ( $x \to -5^+$ ): Approaching $-5$ from values greater than $-5$ ( $x > -5$ ) uses the middle formula piece: $\lim_{x \to -5^+} \sqrt{25 - x^2} = \sqrt{25 - (-5)^2} = \sqrt{25 - 25} = 0$
- **(b) Left-sided limit ( $x \to -5^-)**: Approaching$ -5 $from values less than$ -5 $($ x < -5$) uses the top constant piece: $\lim_{x \to -5^-} (0) = 0$
- (c) Two-sided limit ( $\lim_{x \to -5} f(x)$ ): Because $\lim_{x \to -5^-} f(x) = 0 = \lim_{x \to -5^+} f(x)$ , by the Existence Theorem, the two-sided limit exists: $\lim_{x \to -5} f(x) = 0$
Analyze Boundary $x = 4$ :
- (d) Right-sided limit ( $x \to 4^+$ ): Approaching $4$ from values greater than $4$ ( $x > 4$ ) uses the bottom piece: $\lim_{x \to 4^+} (5x) = 5(4) = 20$
- (e) Left-sided limit ( $x \to 4^-$ ): Approaching $4$ from values less than $4$ ( $x < 4$ ) uses the middle piece: $\lim_{x \to 4^-} \sqrt{25 - x^2} = \sqrt{25 - (4)^2} = \sqrt{25 - 16} = \sqrt{9} = 3$
- (f) Two-sided limit ( $\lim_{x \to 4} f(x)$ ): Since the left-hand limit ( $3$ ) does not equal the right-hand limit ( $20$ ), the two-sided limit does not exist: $\lim_{x \to 4} f(x) = \text{DNE}$ (Classification: Jump Discontinuity)

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, state the Existence Theorem for Two-Sided Limits.
(b) In Worked Example A, explain why the two-sided limit existed at $x = -5$ but failed to exist at $x = 4$ . What type of discontinuity occurred at $x = 4$ ?
(c) Review the linked Engineering Applications section. How do one-sided limits describe the physical response of a circuit breaker or structural material at its failure threshold?

Solution Key:

(a) According to the Conceptual Framework, the Existence Theorem states that $\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L = \lim_{x \to a^+} f(x)$ . Both directional paths must converge to the same finite output.
(b) In Worked Example A, at the boundary $x = -5$ , the left-hand limit $\lim_{x \to -5^-} f(x) = 0$ and the right-hand limit $\lim_{x \to -5^+} f(x) = 0$ are identical, so $\lim_{x \to -5} f(x) = 0$ . At $x = 4$ , the left-hand limit is $3$ and the right-hand limit is $20$ . Because $3 \neq 20$ , the two-sided limit does not exist ( $\text{DNE}$ ), classified as a Jump Discontinuity.
(c) In the linked Engineering Applications (e.g., "Material Tensile Failure"), the left-sided limit models behavior just before fracture ( $80\text{ MPa}$ ), while the right-sided limit models post-fracture behavior ( $0\text{ MPa}$ ). The mismatch indicates a physical jump discontinuity corresponding to total structural snap.

Show your steps and calculations here.

Practice Problem 5

Consider the piecewise-defined function:

g(x) = \begin{cases} 2x + 1 & : x < 1 \\ 6 & : x = 1 \\ x^2 + 2 & : x > 1 \end{cases}

Evaluate the following values: (a) $\lim_{x \to 1^-} g(x)$
(b) $\lim_{x \to 1^+} g(x)$
(c) $\lim_{x \to 1} g(x)$
(d) $g(1)$
(e) Does the value of $g(1)$ affect the value of $\lim_{x \to 1} g(x)$ ? What type of discontinuity exists at $x = 1$ ?

Solution Key:

Left-sided limit ( $x \to 1^-$ ): $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} (2x + 1) = 2(1) + 1 = 3$
Right-sided limit ( $x \to 1^+$ ): $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (x^2 + 2) = (1)^2 + 2 = 3$
Two-sided limit ( $\lim_{x \to 1} g(x)$ ): Since $\lim_{x \to 1^-} g(x) = 3 = \lim_{x \to 1^+} g(x)$ , by the Existence Theorem, the two-sided limit exists and equals $3$ : $\lim_{x \to 1} g(x) = 3$
Function value ( $g(1)$ ): By the piecewise definition, at the single point $x = 1$ , $g(1) = 6$ .
Comparison & Classification: The value $g(1) = 6$ does not affect the limit. Limits evaluate the trend of neighboring points ( $x \neq 1$ ). Because $\lim_{x \to 1} g(x) = 3 \neq g(1) = 6$ , $g(x)$ exhibits a Removable Discontinuity at $x = 1$ .

Show your steps and calculations here.

Practice Problem 6

Evaluate the limit:

\lim_{x \to -1} \frac{x^2 - 3x - 4}{x^2 + 2x - 3}

Solution Key:

Attempt Direct Substitution: Plug $x = -1$ directly into the rational expression:
- Numerator: $(-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0$
- Denominator: $(-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$
Evaluate the Fraction: $\frac{0}{-4} = 0$
Conclusion: Because the denominator is non-zero ( $-4 \neq 0$ ), this is not an indeterminate form ( $\frac{0}{0}$ ). Direct substitution yields the exact limit value via the Quotient Law: $\lim_{x \to -1} \frac{x^2 - 3x - 4}{x^2 + 2x - 3} = 0$

Show your steps and calculations here.

Regroup 2

Discuss how piecewise models describe real-world control systems and physical phase transitions.
Clarify that a zero in the numerator alone evaluates to $0$ , whereas $\frac{0}{0}$ represents a removable discontinuity requiring algebraic simplification.

Skill Block 3

Worked Example A: Factoring Quadratic Expressions

Evaluate the limit:

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}

Worked Solution:

Check Direct Substitution: $\frac{2^2 + 3(2) - 10}{2 - 2} = \frac{4 + 6 - 10}{0} = \frac{0}{0} \quad (\text{Indeterminate})$ Because $x \to 2$ , the factor causing division by zero in both numerator and denominator is $(x - 2)$ .
Factor the Numerator: Using factoring techniques, find factors of $-10$ that sum to $+3$ , which are $+5$ and $-2$ : $x^2 + 3x - 10 = (x + 5)(x - 2)$
Cancel and Evaluate: Cancel $(x - 2)$ : $\lim_{x \to 2} \frac{(x + 5)(x - 2)}{x - 2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7$ (Note: Dividing $(x^2 + 3x - 10)$ directly by $(x - 2)$ via polynomial long division yields the identical quotient $(x + 5)$ with zero remainder).

Worked Example B: Higher-Degree Polynomial Long Division (The $(x - a)$ Factor Trick)

Evaluate the limit:

\lim_{x \to 3} \frac{x^3 - 2x^2 - 4x + 3}{x - 3}

Worked Solution:

Check Direct Substitution:
$\frac{3^3 - 2(3^2) - 4(3) + 3}{3 - 3} = \frac{27 - 18 - 12 + 3}{0} = \frac{0}{0} \quad (\text{Indeterminate})$
Apply the $(x - a)$ Target Factor Trick: Because $x \to 3$ , the factor causing the zero in the cubic numerator is $(x - 3)$ . Instead of struggling with cubic factoring by grouping, use polynomial long division to divide the numerator by $(x - 3)$ .
Perform Polynomial Long Division: Divide $(x^3 - 2x^2 - 4x + 3) \div (x - 3)$ :
- First Term: $x^3 \div x = x^2$ . Multiply: $x^2(x - 3) = x^3 - 3x^2$ . Subtract: $(x^3 - 2x^2) - (x^3 - 3x^2) = x^2$ . Bring down $-4x \implies x^2 - 4x + 3$ .
- Second Term: $x^2 \div x = x$ . Multiply: $x(x - 3) = x^2 - 3x$ . Subtract: $(x^2 - 4x) - (x^2 - 3x) = -x$ . Bring down $+3 \implies -x + 3$ .
- Third Term: $-x \div x = -1$ . Multiply: $-1(x - 3) = -x + 3$ . Subtract: $(-x + 3) - (-x + 3) = 0$ (Remainder $0$ ).
The quotient is $(x^2 + x - 1)$ .
Evaluate Simplified Limit:
$\lim_{x \to 3} \frac{(x - 3)(x^2 + x - 1)}{x - 3} = \lim_{x \to 3} (x^2 + x - 1) = 3^2 + 3 - 1 = 9 + 3 - 1 = 11$

Worked Example C: Radical Conjugate Rationalization

Evaluate the limit:

\lim_{x \to 3} \frac{\sqrt{x^2 + 7} - 4}{x - 3}

Worked Solution:

Check Direct Substitution: $\frac{\sqrt{3^2 + 7} - 4}{3 - 3} = \frac{\sqrt{16} - 4}{0} = \frac{0}{0} \quad (\text{Indeterminate})$ The target zero factor to extract is $(x - 3)$ .
Multiply by the Radical Conjugate: Using radical conjugate rationalization, multiply numerator and denominator by $(\sqrt{x^2 + 7} + 4)$ : $\lim_{x \to 3} \left[ \frac{\sqrt{x^2 + 7} - 4}{x - 3} \cdot \frac{\sqrt{x^2 + 7} + 4}{\sqrt{x^2 + 7} + 4} \right]$
Expand the Numerator & Keep Denominator Factored: $(\sqrt{x^2 + 7} - 4)(\sqrt{x^2 + 7} + 4) = (x^2 + 7) - 16 = x^2 - 9 = (x - 3)(x + 3)$ Rewrite the limit: $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{(x - 3)(\sqrt{x^2 + 7} + 4)}$
Cancel Common Factor and Re-Substitute: Cancel $(x - 3)$ : $\lim_{x \to 3} \frac{x + 3}{\sqrt{x^2 + 7} + 4} = \frac{3 + 3}{\sqrt{3^2 + 7} + 4} = \frac{6}{\sqrt{16} + 4} = \frac{6}{4 + 4} = \frac{6}{8} = \frac{3}{4}$

Worked Example D: Exponential Factoring & Difference of Squares

Evaluate the limit:

\lim_{x \to 0} \frac{e^{2x} - 1}{2e^x - 2}

Worked Solution:

Check Direct Substitution: $\frac{e^0 - 1}{2e^0 - 2} = \frac{1 - 1}{2(1) - 2} = \frac{0}{0} \quad (\text{Indeterminate})$ Here, the zero-inducing term is $(e^x - 1)$ since $e^0 - 1 = 0$ .
Factor Numerator and Denominator:
- Denominator: Factor out constant scalar $2$ : $2e^x - 2 = 2(e^x - 1)$
- Numerator: Treat $e^{2x} = (e^x)^2$ as a difference of squares: $e^{2x} - 1 = (e^x - 1)(e^x + 1)$
Cancel and Evaluate: $\lim_{x \to 0} \frac{(e^x - 1)(e^x + 1)}{2(e^x - 1)} = \lim_{x \to 0} \frac{e^x + 1}{2} = \frac{e^0 + 1}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1$

Worked Example E: Trigonometric Identity Substitution

Evaluate the limit:

\lim_{x \to 0} \frac{\sin^2 x}{1 - \cos x}

Worked Solution:

Check Direct Substitution: $\frac{\sin^2(0)}{1 - \cos(0)} = \frac{0^2}{1 - 1} = \frac{0}{0} \quad (\text{Indeterminate})$ The zero-inducing factor is $(1 - \cos x)$ .
Apply Pythagorean Identity: From Trigonometric Identities, recall $\sin^2 x + \cos^2 x = 1 \implies \sin^2 x = 1 - \cos^2 x$ . Rewrite the numerator as a difference of squares: $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$
Substitute, Cancel, and Evaluate: $\lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x} = \lim_{x \to 0} (1 + \cos x)$ Substitute $x = 0$ : $1 + \cos(0) = 1 + 1 = 2$ (Note: Multiplying numerator and denominator by the trigonometric conjugate $(1 + \cos x)$ yields the identical result).

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Under the Conceptual Framework section, what is the Target Factor Principle? When evaluating a limit as $x \to a$ that gives $\frac{0}{0}$ , what factor are we guaranteed to cancel?
(b) In Worked Example B, how is polynomial long division by $(x - 3)$ used to factor the higher-degree cubic polynomial without guessing? Why is the remainder guaranteed to be zero?
(c) Review the linked Engineering Applications section. How does algebraic resolution prevent divide-by-zero singularities in valve control systems or robot deflection equations?

Solution Key:

(a) The Target Factor Principle states that as $x \to a$ , the term that evaluates to zero is $(x - a)$ . Therefore, whenever direct substitution yields $\frac{0}{0}$ , both the numerator and denominator share the known factor $(x - a)$ .
(b) Because $(x - a)$ is guaranteed to be an exact factor of $P(x)$ , performing polynomial long division of $P(x) \div (x - a)$ eliminates the need for factoring trial-and-error. By the Polynomial Remainder Theorem, the remainder will always be $0$ , and the quotient polynomial is the exact simplified expression ready for evaluation.
(c) In the linked Engineering Applications (e.g. "Flow Control Valve Restriction"), the theoretical formula models velocity as $\frac{x^2 - 5x + 6}{x - 3}$ , which yields $\frac{0}{0}$ at the critical orifice setting $x = 3\text{ cm}$ . Factoring out $(x - 3)$ reveals the continuous physical velocity limit of $1\text{ m/s}$ , preventing division-by-zero crashes in automated valve control algorithms.

Show your steps and calculations here.

Practice Problem 8

Evaluate the limit using factor-and-cancel:

\lim_{x \to 4} \frac{x^2 - 7x + 12}{x - 4}

Solution Key:

Check Direct Substitution:
$\frac{4^2 - 7(4) + 12}{4 - 4} = \frac{16 - 28 + 12}{0} = \frac{0}{0} \quad (\text{Indeterminate})$ The target zero factor is $(x - 4)$ .
Method 1 — Traditional Factoring: Find numbers multiplying to $+12$ and adding to $-7$ , which are $-4$ and $-3$ :
$x^2 - 7x + 12 = (x - 4)(x - 3)$ $\lim_{x \to 4} \frac{(x - 4)(x - 3)}{x - 4} = \lim_{x \to 4} (x - 3) = 4 - 3 = 1$
Method 2 — Polynomial Long Division: Divide $(x^2 - 7x + 12)$ by the known factor $(x - 4)$ :
- $x^2 \div x = x \implies (x^2 - 7x) - (x^2 - 4x) = -3x + 12$
- $-3x \div x = -3 \implies (-3x + 12) - (-3x + 12) = 0$ (Remainder $0$ ).
The quotient is $(x - 3)$ . Substituting $x = 4$ yields $4 - 3 = 1$ .

Show your steps and calculations here.

Practice Problem 9

Evaluate the limit using the $(x - a)$ factor trick with polynomial long division on the cubic denominator:

\lim_{x \to 2} \frac{x^2 + x - 6}{x^3 - x^2 - x - 2}

Solution Key:

Check Direct Substitution:
- Numerator: $2^2 + 2 - 6 = 0$
- Denominator: $2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$ This is an indeterminate form $\frac{0}{0}$ . The target factor to extract from both is $(x - 2)$ .
Factor Numerator:
$x^2 + x - 6 = (x - 2)(x + 3)$
Divide Denominator by $(x - 2)$ via Polynomial Long Division: Divide $(x^3 - x^2 - x - 2) \div (x - 2)$ :
- $x^3 \div x = x^2$ . Multiply: $x^2(x - 2) = x^3 - 2x^2$ . Subtract: $(x^3 - x^2) - (x^3 - 2x^2) = x^2$ . Bring down $-x \implies x^2 - x - 2$ .
- $x^2 \div x = x$ . Multiply: $x(x - 2) = x^2 - 2x$ . Subtract: $(x^2 - x) - (x^2 - 2x) = x$ . Bring down $-2 \implies x - 2$ .
- $x \div x = 1$ . Multiply: $1(x - 2) = x - 2$ . Subtract: $(x - 2) - (x - 2) = 0$ (Remainder $0$ ).
The denominator factors into $(x - 2)(x^2 + x + 1)$ .
Cancel $(x - 2)$ and Evaluate:
$\lim_{x \to 2} \frac{(x - 2)(x + 3)}{(x - 2)(x^2 + x + 1)} = \lim_{x \to 2} \frac{x + 3}{x^2 + x + 1} = \frac{2 + 3}{2^2 + 2 + 1} = \frac{5}{4 + 2 + 1} = \frac{5}{7}$

Show your steps and calculations here.

Practice Problem 10

Evaluate the limit using both conjugate rationalization and factoring:

\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1}

Solution Key:

Check Direct Substitution: $\frac{1 - 1}{\sqrt{1} - 1} = \frac{0}{0} \quad (\text{Indeterminate})$
Method 1 — Radical Conjugate Rationalization: Multiply numerator and denominator by $(\sqrt{x} + 1)$ : $\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \lim_{x \to 1} \frac{(x - 1)(\sqrt{x} + 1)}{(\sqrt{x})^2 - 1^2} = \lim_{x \to 1} \frac{(x - 1)(\sqrt{x} + 1)}{x - 1}$ Cancel $(x - 1)$ : $\lim_{x \to 1} (\sqrt{x} + 1) = \sqrt{1} + 1 = 2$
Method 2 — Difference of Squares Factoring: Factor the numerator as $(\sqrt{x} - 1)(\sqrt{x} + 1)$ : $\lim_{x \to 1} \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} - 1} = \lim_{x \to 1} (\sqrt{x} + 1) = 2$

Show your steps and calculations here.

Practice Problem 11

Evaluate the limit using a trigonometric identity:

\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^2 x}{2\cos x}

Solution Key:

Check Direct Substitution: $\frac{1 - \sin^2(\pi/2)}{2\cos(\pi/2)} = \frac{1 - (1)^2}{2(0)} = \frac{0}{0} \quad (\text{Indeterminate})$
Apply Pythagorean Identity: Substitute $1 - \sin^2 x = \cos^2 x$ into the numerator: $\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{2\cos x}$
Cancel Common Factor and Evaluate: Cancel one factor of $\cos x$ from numerator and denominator: $\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2} = \frac{\cos(\pi/2)}{2} = \frac{0}{2} = 0$

Show your steps and calculations here.

Regroup 3

Summarize the three algebraic tools for resolving $\frac{0}{0}$ :
1. Factor and Cancel (using traditional factoring or polynomial long division by $(x - a)$ ).
2. Radical Conjugate Rationalization (multiplying by $\frac{\sqrt{A}+B}{\sqrt{A}+B}$ ).
3. Trigonometric Identities (using Pythagorean and conjugate substitutions).

Synthesis Wrap-up

Core takeaways from Session 03:
1. Limit Laws: Valid only when component limits exist as finite numbers; allow distributive algebraic evaluation (Sum, Difference, Constant Multiple, Product, Quotient, and Power & Root Laws).
2. Piecewise Functions: Evaluated at boundary thresholds using one-sided limit strategies and verifying agreement via the Existence Theorem.
3. Indeterminate Forms ( $\frac{0}{0}$ ): Signify removable discontinuities (holes) where the zero factor $(x - a)$ is resolved via factoring or polynomial long division, radical conjugates, or trigonometric identities before re-substituting.
Connections to Knewton Alta Homework: Computing Limits with Algebraic Techniques.

MATH-181 (Calculus I) Curriculum

Session 03 - Computing Limits with Algebraic Techniques

Micro-Lecture

Skill Block 1

Worked Example A: Evaluating Combined Functions Using Limit Laws

Practice Problem 1: Concept & Reading Check

Practice Problem 2

Practice Problem 3

Regroup 1

Skill Block 2

Worked Example A: Evaluating Limits of Piecewise Functions at Boundary Points

Practice Problem 4: Concept & Reading Check

Practice Problem 5

Practice Problem 6

Regroup 2

Skill Block 3

Worked Example A: Factoring Quadratic Expressions

Worked Example B: Higher-Degree Polynomial Long Division (The (x−a)(x - a)(x−a) Factor Trick)

Worked Example C: Radical Conjugate Rationalization

Worked Example D: Exponential Factoring & Difference of Squares

Worked Example E: Trigonometric Identity Substitution

Practice Problem 7: Concept & Reading Check

Practice Problem 8

Practice Problem 9

Practice Problem 10

Practice Problem 11

Regroup 3

Synthesis Wrap-up

Worked Example B: Higher-Degree Polynomial Long Division (The $(x - a)$ Factor Trick)