Session 03 - Computing Limits with Algebraic Techniques

📅 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.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

⚡

Electrical Engineering

Electrical Circuit Mesh Network (Electrical Engineering)

An electrical hardware engineer is monitoring three separate fluctuating voltage branches ( $V_1$ , $V_2$ , and $V_3$ ) inside a microgrid inverter circuit as the system reaches a steady initialization threshold ( $t \to s$ ). Laboratory telemetry provides the following individual component limits:

$\lim_{t\to s} V_1(t) = 12\text{ V}$
$\lim_{t\to s} V_2(t) = 5\text{ V}$
$\lim_{t\to s} V_3(t) = 3\text{ V}$

The net safety voltage profile of the circuit loop is modeled by the function $V_{\text{net}}(t) = V_1(t) + V_2(t) - V_3(t)$ .

State which Limit Laws allow you to evaluate the limit of $V_{\text{net}}(t)$ as $t \to s$ .
Compute the steady-state initialization limit $\lim_{t\to s} V_{\text{net}}(t)$ .

💡 Show Worked Derivation

Part 1: Limit Laws Applied

Because the individual limits $\lim_{t\to s} V_1(t)$ , $\lim_{t\to s} V_2(t)$ , and $\lim_{t\to s} V_3(t)$ all exist and are finite real numbers, we can apply the Sum Law and the Difference Law to distribute the limit across each voltage term independently:

\lim_{t\to s} V_{\text{net}}(t) = \lim_{t\to s} [V_1(t) + V_2(t) - V_3(t)] = \lim_{t\to s} V_1(t) + \lim_{t\to s} V_2(t) - \lim_{t\to s} V_3(t)

Part 2: Calculation

Substitute the given sensor limits into the distributed expression:

\lim_{t\to s} V_{\text{net}}(t) = 12\text{ V} + 5\text{ V} - 3\text{ V} = 14\text{ V}

Conclusion: The net steady-state initialization limit for the circuit loop is $14\text{ V}$ . In electrical engineering, this confirms that the combined branch voltages stabilize to a predictable operating voltage without inductive spikes or brownouts.

🚀

Aerospace Engineering

Rocket Payload Velocity (Mechanical & Aerospace Engineering)

An aerospace engineer is analyzing the launch trajectory of a multi-stage sounding rocket during atmospheric ascent. The instantaneous velocity $V(t)$ (in $\text{m/s}$ ) depends on the engine thrust factor $T(t)$ (in $\text{m/s}^2$ ) and the normalized propellant mass burn factor $M(t)$ relative to burn-time $t$ (in seconds), governed by $V(t) = T(t) \cdot M(t)$ .

Telemetry data confirms the following limits as the rocket approaches the end of first-stage burn at $t \to 4\text{ seconds}$ :

$\lim_{t\to 4} T(t) = 8\text{ m/s}^2$
$\lim_{t\to 4} M(t) = 12$

Identify the Limit Law needed to evaluate $\lim_{t\to 4} V(t)$ .
Calculate the payload velocity limit $\lim_{t\to 4} V(t)$ .

💡 Show Worked Derivation

Part 1: Limit Law Applied

Since both $\lim_{t\to 4} T(t)$ and $\lim_{t\to 4} M(t)$ exist as finite values, we apply the Product Law for Limits, which states that the limit of a product is the product of the limits:

\lim_{t\to 4} V(t) = \lim_{t\to 4} [T(t) \cdot M(t)] = \left(\lim_{t\to 4} T(t)\right) \cdot \left(\lim_{t\to 4} M(t)\right)

Part 2: Calculation

Substitute the given telemetry limits:

\lim_{t\to 4} V(t) = (8\text{ m/s}^2) \cdot (12) = 96\text{ m/s}

Conclusion: The expected payload velocity as the rocket approaches the 4-second stage transition is $96\text{ m/s}$ . The Product Law allows aerospace engineers to decouple complex multi-variable propulsion dynamics into distinct, independently tested subsystem limits.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Applying Fundamental Limit Laws

General Overview
Skill Block 1 establishes the procedural foundation for decomposing complex mathematical operations using standard Limit Laws without relying on graphs or numerical tables.

Conceptual Framework

Distributive Nature of Limits: If the component limits exist as finite real numbers ( $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ ), the limit operator distributes across addition, subtraction, multiplication, division, and exponents.
The Finite Existence Constraint: Limit laws only apply when each constituent limit evaluates to a determinate, finite real number. If an individual limit yields $\pm\infty$ or an indeterminate form ( $\frac{0}{0}$ ), the expression cannot be split until it is algebraically simplified.
Non-Zero Denominator Rule: The Quotient Law requires the denominator's limit to be non-zero ( $\lim_{x \to a} g(x) \neq 0$ ). If the denominator limit is zero, direct quotient distribution is invalid.

Key Equations & Structural Rules

Sum Law: $\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L + M$
Difference Law: $\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) = L - M$
Constant Multiple Law: $\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) = c \cdot L \quad (\text{where } c \text{ is constant})$
Product Law: $\lim_{x \to a} [f(x) \cdot g(x)] = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right) = L \cdot M$
Quotient Law: $\lim_{x \to a} \left[\frac{f(x)}{g(x)}\right] = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{L}{M} \quad (\text{provided } M \neq 0)$
Power & Root Law: $\lim_{x \to a} [f(x)]^{n/m} = \left[\lim_{x \to a} f(x)\right]^{n/m} = L^{n/m} \quad (\text{where } n, m \text{ are integers, } m \neq 0)$

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) $\lim_{x \to 1} \frac{g(x)}{h(x)}$

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Check the Conceptual Framework regarding finite limits, review step 4 of Worked Example A, and refer to the linked Engineering Application <a class="text-accent underline font-semibold" href="/garden/electrical-circuit-mesh-network--electrical-engineering-">Electrical Circuit Mesh Network (Electrical Engineering)</a>.

Answer: * (a) Limit laws apply only when individual component limits exist as finite real numbers. They cannot be applied directly if any component yields $\pm\infty$ or an indeterminate form like $\frac{0}{0}$ .

(b) The limit was moved inside the square bracket via the Power Law, and the constant coefficient $3$ was factored out via the Constant Multiple Law, yielding $[3(2)]^2 = 36$ .
(c) The Sum and Difference Laws allow distributing the limit operator across individual branch voltage functions ( $V_1(t) + V_2(t) - V_3(t)$ ) to find the net steady-state voltage.

Step-by-Step Execution

(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.

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 $\lim_{x \to 2} \frac{f(x)}{h(x)}$ ?

Practice Space Solve this problem on paper or tablet.

For (a), distribute using Constant Multiple and Difference Laws: $2(4) - 3(-2)$ .
For (b), evaluate numerator and denominator separately using Product and Sum Laws. Check that the denominator limit is non-zero.
For (c), move the limit inside the radical using the Power & Root Law.
For (d), check the denominator limit value $\lim_{x \to 2} h(x)$ against the requirements of the Quotient Law.

Answer: - (a) $14$

(b) $-4$
(c) $3$
(d) The Quotient Law requires the denominator's limit to be non-zero, but $\lim_{x \to 2} h(x) = 0$ .

Step-by-Step Execution

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.

Practice Problem 3

Evaluate the limit:

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

Practice Space Solve this problem on paper or tablet.

Apply the Power & Root Law to bring the limit inside the fractional power: $\left[\lim_{x \to 0} (x - 8)\right]^{2/3}$ . Then evaluate the base and simplify the exponent $(-8)^{2/3} = (\sqrt[3]{-8})^2$ .

Answer:

4

Step-by-Step Execution

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: $(-8)^{2/3} = \left(\sqrt[3]{-8}\right)^2 = (-2)^2 = 4$

Segment 2 of 8

📅 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 $\pm\infty$ .

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

⚡

Electrical Engineering

Electrical Surge Protection Switch (Computer & Electrical Engineering)

A computer hardware engineer designs an automated over-voltage surge protection circuit for an embedded server power supply. The electric current profile $I(t)$ (in amperes) relative to initialization time $t$ (in milliseconds) exhibits a rapid switching threshold at $t = 2\text{ ms}$ :

I(t) = \begin{cases} 4 & : t < 2 \\ t^2 + 5 & : t \ge 2 \end{cases}

Evaluate the left-sided current limit $\lim_{t\to 2^-} I(t)$ .
Evaluate the right-sided current limit $\lim_{t\to 2^+} I(t)$ .
Determine whether the overall limit $\lim_{t\to 2} I(t)$ exists, and explain what the result indicates about the electrical power transition.

💡 Show Worked Derivation

Step-by-Step Solution

Left-sided limit ( $t \to 2^-$ ): For times prior to the 2 ms switching threshold ( $t < 2$ ), the circuit operates in steady baseline mode governed by $I(t) = 4$ :
$\lim_{t\to 2^-} I(t) = \lim_{t\to 2^-}(4) = 4\text{ A}$
Right-sided limit ( $t \to 2^+$ ): For times at or after the switching event ( $t \ge 2$ ), the circuit enters the active surge suppression state governed by $I(t) = t^2 + 5$ :
$\lim_{t\to 2^+} I(t) = \lim_{t\to 2^+}(t^2 + 5) = 2^2 + 5 = 4 + 5 = 9\text{ A}$
Two-sided limit evaluation and physical interpretation: Comparing the directional limits:
$\lim_{t\to 2^-} I(t) = 4\text{ A} \quad \neq \quad \lim_{t\to 2^+} I(t) = 9\text{ A}$ Since the left-hand and right-hand limits are not equal, the overall two-sided limit Does Not Exist (DNE).

Engineering Implication: This limit failure is a jump discontinuity with a sudden step increase of $5\text{ amperes}$ at $t = 2\text{ ms}$ . This indicates a sharp switching transient, alerting the computer engineer to incorporate decoupling capacitors to suppress switching noise and protect sensitive logic components.

🏗️

Civil Engineering

Material Tensile Failure (Mechanical & Civil Engineering)

A materials testing laboratory subjects a high-tensile steel structural reinforcement rod to axial tensile loading. The internal tensile stress $S(x)$ (in megapascals, $\text{MPa}$ ) is measured as a function of elongation displacement $x$ (in millimeters), up to its critical physical fracture threshold at $x = 10\text{ mm}$ :

S(x) = \begin{cases} 8x & : 0 \le x < 10 \\ 0 & : x \ge 10 \quad \text{(The rod fractures completely)} \end{cases}

Evaluate the left-sided limit $\lim_{x\to 10^-} S(x)$ .
Evaluate the right-sided limit $\lim_{x\to 10^+} S(x)$ .
Determine whether $\lim_{x\to 10} S(x)$ exists, and interpret the structural significance of the boundary behavior.

💡 Show Worked Derivation

Step-by-Step Solution

Left-sided limit ( $x \to 10^-$ ): Approaching from the elastic/plastic deformation zone prior to fracture ( $x < 10$ ), the stress climbs linearly:
$\lim_{x\to 10^-} S(x) = \lim_{x\to 10^-}(8x) = 8(10) = 80\text{ MPa}$
Right-sided limit ( $x \to 10^+$ ): Approaching from the post-fracture region ( $x \ge 10$ ), the severed rod can bear no load:
$\lim_{x\to 10^+} S(x) = \lim_{x\to 10^+}(0) = 0\text{ MPa}$
Limit existence and structural interpretation: Comparing the one-sided limits:
$\lim_{x\to 10^-} S(x) = 80\text{ MPa} \quad \neq \quad \lim_{x\to 10^+} S(x) = 0\text{ MPa}$ Because the one-sided limits do not match, the two-sided limit Does Not Exist (DNE).

Engineering Implication: The failure of the limit corresponds to an instantaneous drop in load capacity from $80\text{ MPa}$ to $0\text{ MPa}$ at the fracture point $x = 10\text{ mm}$ . In civil and structural engineering, identifying such jump discontinuities is essential for defining safe allowable elongation limits and safety factors before ultimate tensile failure occurs.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Evaluating Common Limits and One-Sided Piecewise Limits

General Overview
Skill Block 2 establishes core proficiency in evaluating fundamental limits of constant functions and the identity function, and analyzing piecewise-defined functions at their boundary thresholds using directional one-sided limit strategies.

Conceptual Framework

Trivial Limits (Constants and Identity): The limit of a constant function is simply the constant itself ( $\lim_{x \to a} c = c$ ), and the limit of the identity function is the approach coordinate ( $\lim_{x \to a} x = a$ ).
The Existence Theorem for Two-Sided Limits: By the Existence Theorem for Two-Sided Limits, an overall limit $\lim_{x \to a} f(x) = L$ exists if and only if both one-sided limits exist and agree: $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$ .
Piecewise Boundary Transitions: Piecewise functions use different formulas across distinct intervals. To determine if an overall limit exists at a domain boundary $x = c$ , you must apply strategies for one-sided limits: evaluate $\lim_{x \to c^-} f(x)$ using the left-hand formula and $\lim_{x \to c^+} f(x)$ using the right-hand formula.
Diagnosing Discontinuities: If the directional limits exist but disagree, the boundary exhibits a jump discontinuity. If values approach infinity, it exhibits an infinite discontinuity.

Key Equations & Structural Rules

Constant Rule: $\lim_{x \to a} c = c$
Identity Rule: $\lim_{x \to a} x = a$
Existence Theorem: $\lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L \iff \lim_{x \to a} f(x) = L$
Discontinuity Condition: If $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$ (both finite), then $\lim_{x \to a} f(x) = \text{DNE}$ (see Jump Discontinuity).

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) $\lim_{x \to 4} f(x)$

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Look at the Existence Theorem under Conceptual Framework, review the solution steps for Worked Example A, and examine the linked Engineering Application <a class="text-accent underline font-semibold" href="/garden/material-tensile-failure--mechanical---civil-engineering-">Material Tensile Failure (Mechanical & Civil Engineering)</a>.

Answer: * (a) A two-sided limit exists if and only if both the left-hand and right-hand limits exist and are equal: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$ .

(b) At $x = -5$ , both one-sided limits equaled $0$ , so the limit exists. At $x = 4$ , the left-hand limit was $3$ while the right-hand limit was $20$ ; since $3 \neq 20$ , the limit fails to exist due to a Jump Discontinuity.
(c) One-sided limits describe the sudden jump in state (e.g. stress dropping from $80\text{ MPa}$ to $0\text{ MPa}$ when a rod fractures, or current jumping across a surge switch).

Step-by-Step Execution

(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.

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 $x = 1$ ?

Practice Space Solve this problem on paper or tablet.

For (a), use one-sided limit strategies for $x < 1$ : $2(1) + 1$ .
For (b), use the piece for $x > 1$ : $(1)^2 + 2$ .
For (c), check if the left-hand and right-hand limits match using the Existence Theorem.
For (d), read the value defined specifically at $x = 1$ .
For (e), recall the fundamental distinction between approaching behavior and point evaluation, and see Removable Discontinuity.

Answer: - (a) $3$

(b) $3$
(c) $3$
(d) $6$
(e) No. A limit depends strictly on approaching behavior ( $x \to 1$ ). $g(x)$ has a Removable Discontinuity (Hole) at $(1, 3)$ with an isolated point at $(1, 6)$ .

Step-by-Step Execution

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 $x = 1$ .

Practice Problem 6

Evaluate the limit:

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

Practice Space Solve this problem on paper or tablet.

Always test direct substitution first. Plug $x = -1$ into the numerator and denominator. Is the denominator zero?

Answer:

0

Step-by-Step Execution

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: $\lim_{x \to -1} \frac{x^2 - 3x - 4}{x^2 + 2x - 3} = 0$

Segment 4 of 8

📅 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.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🏗️

Civil Engineering

Flow Control Valve Restriction (Civil & Environmental Engineering)

A water resources engineer designs an automated municipal overflow valve for a storm runoff conduit. The flow discharge velocity factor $V(x)$ (in $\text{m/s}$ ) through the throttle orifice is modeled as a function of the aperture opening width $x$ (in centimeters) by:

V(x) = \frac{x^2 - 5x + 6}{x - 3}

During valve tuning, the aperture is calibrated around the critical transition setting $x = 3\text{ cm}$ .

Explain why direct evaluation at $x = 3\text{ cm}$ fails.
Apply an algebraic factoring strategy to find the limiting flow velocity $\lim_{x\to 3} V(x)$ .

💡 Show Worked Derivation

Step-by-Step Solution

Direct substitution check: Plugging $x = 3$ directly into the velocity function:
$V(3) = \frac{3^2 - 5(3) + 6}{3 - 3} = \frac{9 - 15 + 6}{0} = \frac{0}{0}$ Direct evaluation yields the indeterminate form $\frac{0}{0}$ . The function value is undefined at $x = 3$ , meaning the raw formula experiences a removable singularity at that exact aperture width.
Algebraic factoring and limit evaluation: Factor the quadratic polynomial in the numerator by finding factors of $+6$ that sum to $-5$ , which are $-2$ and $-3$ :
$x^2 - 5x + 6 = (x - 2)(x - 3)$

Rewrite the limit expression:
$\lim_{x\to 3} V(x) = \lim_{x\to 3} \frac{(x - 2)(x - 3)}{x - 3}$

Since $x \to 3$ implies $x \neq 3$ , cancel the common zero-inducing factor $(x - 3)$ :
$\lim_{x\to 3} V(x) = \lim_{x\to 3} (x - 2)$

Apply direct substitution to the simplified expression:
$3 - 2 = 1\text{ m/s}$

Conclusion: The limiting flow velocity as the aperture width approaches $3\text{ cm}$ is $1\text{ m/s}$ . Factoring eliminates the mathematical singularity, allowing the civil/environmental engineer to determine the continuous flow rate through the valve without division-by-zero errors.

⚙️

Mechanical Engineering

Robot Arm Pivot Deflection (Entertainment & Mechanical Engineering)

A mechatronics engineer designing an animatronic stage prop for an entertainment production is evaluating the transient dynamic tip deflection $D(t)$ (in millimeters) of a robotic arm during a sudden high-speed rotation. The deflection is modeled as a function of elapsed time $t$ (in seconds) by:

D(t) = \frac{\sqrt{t + 7} - 3}{t - 2}

The motion controller requires the exact theoretical deflection limit at the pivot transition timestamp $t = 2\text{ seconds}$ .

Verify that direct substitution at $t = 2$ produces an indeterminate form.
Use radical conjugate rationalization to compute $\lim_{t\to 2} D(t)$ .

💡 Show Worked Derivation

Step-by-Step Solution

Direct substitution check: Substitute $t = 2$ directly into the deflection formula:
$D(2) = \frac{\sqrt{2 + 7} - 3}{2 - 2} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$ Direct substitution yields the indeterminate form $\frac{0}{0}$ , meaning the deflection value cannot be computed without resolving the removable discontinuity.
Radical conjugate rationalization: Multiply both the numerator and denominator by the conjugate of the numerator, $(\sqrt{t + 7} + 3)$ :
$\lim_{t\to 2} D(t) = \lim_{t\to 2} \left[ \frac{\sqrt{t + 7} - 3}{t - 2} \cdot \frac{\sqrt{t + 7} + 3}{\sqrt{t + 7} + 3} \right]$

Expand the numerator using the difference of squares identity $(a - b)(a + b) = a^2 - b^2$ :
$(\sqrt{t + 7} - 3)(\sqrt{t + 7} + 3) = (\sqrt{t + 7})^2 - 3^2 = (t + 7) - 9 = t - 2$

Rewrite the limit with the expanded numerator and factored denominator:
$\lim_{t\to 2} \frac{t - 2}{(t - 2)(\sqrt{t + 7} + 3)}$

Cancel the common factor $(t - 2)$ for $t \neq 2$ :
$\lim_{t\to 2} \frac{1}{\sqrt{t + 7} + 3}$

Evaluate by substituting $t = 2$ :
$\frac{1}{\sqrt{2 + 7} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}\text{ mm} \approx 0.167\text{ mm}$

Conclusion: The instantaneous deflection of the robotic arm at the transition point is $\frac{1}{6}\text{ mm}$ ( $\approx 0.167\text{ mm}$ ). Rationalizing the conjugate removes the numerical singularity, providing precise structural deflection data for programming stage automation safety limits.

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Analytical and Algebraic Techniques for Indeterminate Forms

General Overview
Skill Block 3 presents the systematic analytical workflow for resolving limits when direct substitution yields the indeterminate form $\frac{0}{0}$ . Students learn to apply three primary algebraic tools: Factor and Cancel (incorporating both traditional factoring and polynomial long division), Radical Conjugate Rationalization, and Trigonometric Identities (grounded in core trigonometric relationships).

Conceptual Framework

The Indeterminate Form $\frac{0}{0}$ : Direct substitution yielding $\frac{0}{0}$ indicates a removable discontinuity (a hole in the graph) at $x = a$ .
The Target Factor Principle: When evaluating $\lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{0}{0}$ , we always know the exact factor causing division by zero: it is $(x - a)$ , because $(a - a) = 0$ . Every algebraic technique aims to expose and cancel this specific factor.
The Polynomial Long Division Shortcut: Traditional factoring is reliable for simple quadratics, but for complex or higher-degree polynomials in the numerator or denominator, guessing factors is unnecessary. Since $x = a$ is a root, you can perform polynomial long division of the polynomial directly by $(x - a)$ . The division is mathematically guaranteed to leave a remainder of $0$ , yielding the simplified quotient directly!
Radical Conjugates: Expressions containing square roots like $(\sqrt{A} - B)$ are resolved using conjugate rationalization by multiplying by $(\sqrt{A} + B)$ . The difference of squares $(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2$ frees the variable from the radical to expose $(x - a)$ .
Trigonometric Substitutions: For trigonometric expressions yielding $\frac{0}{0}$ , we apply trigonometric identities (such as the Pythagorean identity $\sin^2 x = 1 - \cos^2 x = (1-\cos x)(1+\cos x)$ from Trigonometric Identities) to factor out the zero term.

Key Equations & Structural Rules

Target Factor: $(x - a)$ is always the common factor canceled when $x \to a$ .
Difference of Squares: $a^2 - b^2 = (a - b)(a + b)$
Conjugate Identity: $(\sqrt{A} - B)(\sqrt{A} + B) = A - B^2$
Pythagorean Identity: $\sin^2 x + \cos^2 x = 1 \implies \sin^2 x = 1 - \cos^2 x \quad \text{and} \quad \cos^2 x = 1 - \sin^2 x$
Trig Difference of Squares: $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$

Worked Example A: Factoring Quadratic Expressions

Evaluate the limit:

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

💡 Reveal 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}

💡 Reveal 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}

💡 Reveal 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}

💡 Reveal 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}

💡 Reveal 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?

Practice Space Solve this problem on paper or tablet.

Review the Target Factor Principle and long division shortcut under Conceptual Framework, and review the steps of Worked Example B.

Answer: * (a) The factor causing $\frac{0}{0}$ is always $(x - a)$ .

(b) Dividing the polynomial by $(x - a)$ produces the quotient directly; the remainder is always zero because $x = a$ is an exact root.
(c) It eliminates mathematical division-by-zero singularities in physical equations, allowing engineers to determine continuous operating values (e.g. flow rates and deflection limits).

Step-by-Step Execution

(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.

Practice Problem 8

Evaluate the limit using factor-and-cancel:

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

Practice Space Solve this problem on paper or tablet.

Check direct substitution first. Since $x \to 4$ , the known zero factor to cancel is $(x - 4)$ . Factor the quadratic numerator or divide $(x^2 - 7x + 12)$ by $(x - 4)$ via polynomial long division.

Answer:

1

Step-by-Step Execution

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 $4 - 3 = 1$ .

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}

Practice Space Solve this problem on paper or tablet.

Direct substitution gives $\frac{0}{0}$ . Since $x \to 2$ , both numerator and denominator contain the factor $(x - 2)$ . Factor the numerator quadratic directly, and divide the cubic denominator by $(x - 2)$ via polynomial long division.

Answer:

\frac{5}{7}

Step-by-Step Execution

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:
$\lim_{x \to 2} \frac{(x - 2)(x + 3)}{(x - 2)(x^2 + x + 1)} = \lim_{x \to 2} \frac{x + 3}{x^2 + x + 1} = \frac{2 + 3}{2^2 + 2 + 1} = \frac{5}{4 + 2 + 1} = \frac{5}{7}$

Practice Problem 10

Evaluate the limit using both conjugate rationalization and factoring:

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

Practice Space Solve this problem on paper or tablet.

Direct substitution gives $\frac{0}{0}$ . The target zero factor to cancel is $(x - 1)$ or $(\sqrt{x} - 1)$ . Multiply by the conjugate $(\sqrt{x} + 1)$ or factor the numerator as $(x - 1) = (\sqrt{x} - 1)(\sqrt{x} + 1)$ .

Answer:

2

Step-by-Step Execution

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 $(\sqrt{x} - 1)(\sqrt{x} + 1)$ : $\lim_{x \to 1} \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} - 1} = \lim_{x \to 1} (\sqrt{x} + 1) = 2$

Practice Problem 11

Evaluate the limit using a trigonometric identity:

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

Practice Space Solve this problem on paper or tablet.

Direct substitution gives $\frac{1 - 1}{2(0)} = \frac{0}{0}$ . Apply the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ from Trigonometric Identities in the numerator, then cancel a common factor of $\cos x$ .

Answer:

0

Step-by-Step Execution

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: $\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2} = \frac{\cos(\pi/2)}{2} = \frac{0}{2} = 0$

Segment 6 of 8

📅 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).

Segment 7 of 8

📅 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.

Segment 8 of 8

Computing Limits with Algebraic Techniques

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Electrical Circuit Mesh Network (Electrical Engineering)

Rocket Payload Velocity (Mechanical & Aerospace Engineering)

Worked Example A: Evaluating Combined Functions Using Limit Laws

Practice Problem 1: Concept & Reading Check

Step-by-Step Execution

Practice Problem 2

Step-by-Step Execution

Practice Problem 3

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Electrical Surge Protection Switch (Computer & Electrical Engineering)

Material Tensile Failure (Mechanical & Civil Engineering)

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

Practice Problem 4: Concept & Reading Check

Step-by-Step Execution

Practice Problem 5

Step-by-Step Execution

Practice Problem 6

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

Flow Control Valve Restriction (Civil & Environmental Engineering)

Robot Arm Pivot Deflection (Entertainment & Mechanical Engineering)

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

Step-by-Step Execution

Practice Problem 8

Step-by-Step Execution

Practice Problem 9

Step-by-Step Execution

Practice Problem 10

Step-by-Step Execution

Practice Problem 11

Step-by-Step Execution

📅 Regroup 3

📅 Synthesis Wrap-up

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