Session 04 - Infinite Limits and Limits at Infinity

📅 Micro-Lecture

1. Infinite Limits & Physical Singularities

Mathematical Foundation: When direct substitution into a simplified quotient produces $\frac{c}{0}$ ( $c \neq 0$ ), the function grows without bound ( $\pm\infty$ ). The vertical line $x = a$ is a Vertical Asymptote, representing an Infinite Discontinuity. We determine the direction ( $+\infty$ vs $-\infty$ ) using directional signed zero analysis ( $0^+$ vs $0^-$ ).
Engineering Context: Infinite limits characterize localized physical singularities, such as mechanical resonance in civil infrastructure (unbounded vibration amplitudes when external driving frequencies match structural natural frequencies) or electrical dielectric breakdown and thermal runaway in high-voltage insulators.

2. Limits at Infinity & Global Growth Dominance

Mathematical Foundation: A Limit at Infinity ( $\lim_{x \to \pm\infty} f(x)$ ) analyzes the global end behavior of a function. By the principle of Growth Dominance, higher-power leading terms completely dominate lower-order terms as $|x| \to \infty$ . When this limit converges to a finite value $L$ , the line $y = L$ is a Horizontal Asymptote.
Engineering Context: Limits at infinity define physical ceilings, saturation capacities, and steady-state thresholds, including the maximum product throughput ( $V_{\max}$ ) in chemical bioreactors and asymptotic response latencies in distributed computer cloud networks.

3. Advanced Asymptotic Forms: Radical Conjugates & Exponential Ceilings

Mathematical Foundation: Forms like $\infty - \infty$ at infinity are indeterminate and require radical conjugate multiplication to convert into determinate ratios. When evaluating limits as $x \to -\infty$ containing square roots, the absolute value rule $\sqrt{x^2} = |x| = -x$ introduces a critical negative sign.
Engineering Context: These methods allow aerospace engineers to compute the steady-state terminal velocity of re-entry vehicles under aerodynamic drag and enable computer engineers to program calibration offset margins for autonomous vehicle radar tracking.

Segment 1 of 8

Skill Block 1

Real-Life Engineering Applications

🏗️

Civil Engineering

Structural Resonance Threshold (Civil Engineering)

A structural engineer modeling dynamic wind shear on a suspended pedestrian walkway determines that the structural vibration amplitude $A(\omega)$ (in millimeters) as a function of the wind vortex shedding frequency $\omega$ (in radians per second) is given by:

A(\omega) = \frac{15}{64 - \omega^2}

The walkway structure possesses an un-damped natural resonance frequency at $\omega_0 = 8\text{ rad/s}$ . To size dynamic tuned mass dampers, the engineering team must determine how the vibration amplitude escalates as the wind excitation frequency increases from lower values and approaches this critical resonance threshold.

Classify the behavior of $A(\omega)$ as $\omega \to 8^-$ using directional sign analysis.
Evaluate the limit analytically: $\lim_{\omega \to 8^{-}} \frac{15}{64 - \omega^2}$
State the physical significance of this mathematical behavior for structural integrity.

💡 Show Worked Derivation

Part 1: Directional Approach & Sign Analysis

As $\omega \to 8^-$ , the excitation frequency approaches $8\text{ rad/s}$ from below (e.g., $\omega = 7.99\text{ rad/s}$ ).

Numerator: The constant value is $+15 > 0$ .
Denominator: Because $\omega < 8$ , we have $\omega^2 < 64$ , which means $64 - \omega^2 > 0$ . The denominator approaches $0$ through strictly positive values ( $0^+$ ).

Part 2: Analytical Limit Calculation

Using directional zero sign analysis:

\lim_{\omega \to 8^{-}} \frac{15}{64 - \omega^2} = \frac{+15}{0^+} = +\infty

Part 3: Physical Engineering Significance

The limit diverging to $+\infty$ indicates that without mechanical damping, structural vibration amplitudes grow without bound near the natural frequency $\omega_0 = 8\text{ rad/s}$ . The vertical asymptote at $\omega = 8$ represents a catastrophic mechanical resonance boundary where cyclical wind energy pumps unbounded kinetic energy into the structure, necessitating tuned mass dampers to suppress resonant failure.

⚡

Electrical Engineering

Thermal Runaway in Electrical Discharges (Electrical Engineering)

In high-voltage transmission substation equipment, the leakage current intensity $I$ (in milliamperes) across a deteriorating ceramic insulator gap is modeled as a function of the applied breakdown electrical stress $V$ (in kilovolts) by the rational expression:

I(V) = \frac{-2.5}{V - 12}

If localized voltage surges approach the critical breakdown dielectric rating of $V_c = 12\text{ kV}$ from an overvoltage condition ( $V \to 12^+$ ), substation protection engineers need to map the polarity and magnitude trajectory of the energy discharge.

Classify the behavior of $I(V)$ as $V \to 12^+$ using directional sign analysis.
Evaluate the limit analytically: $\lim_{V \to 12^{+}} \frac{-2.5}{V - 12}$
Explain how the resulting vertical asymptote informs arc-flash circuit breaker trigger ratings.

💡 Show Worked Derivation

Part 1: Directional Approach & Sign Analysis

As $V \to 12^+$ , the voltage stress approaches $12\text{ kV}$ from above ( $V > 12$ , e.g., $V = 12.001\text{ kV}$ ).

Numerator: The constant value is $-2.5 < 0$ (negative).
Denominator: Since $V > 12$ , $V - 12 > 0$ . The denominator approaches $0$ through positive values ( $0^+$ ).

Part 2: Analytical Limit Calculation

Applying the signed zero rule:

\lim_{V \to 12^{+}} \frac{-2.5}{V - 12} = \frac{-2.5}{0^+} = -\infty

Part 3: Engineering Significance

The unbounded negative surge ( $-\infty$ ) models rapid reverse polarity ionization and thermal avalanche breakdown across the insulator. In substation engineering, this vertical asymptote defines the irreversible dielectric failure boundary; protection relays must trigger solid-state arc-flash breakers well before the system operates within the asymptotic neighborhood of $V = 12\text{ kV}$ .

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

Infinite Limits and Vertical Asymptotes (Directional Zero Sign Analysis)

General Overview
Skill Block 1 develops analytical methods to evaluate unbounded function behavior near points of zero denominator division and to distinguish true vertical asymptotes from removable discontinuities (holes).

Conceptual Framework

Unbounded Growth vs Finite Limit: Writing $\lim_{x \to a} f(x) = \infty$ does not mean the limit exists as a real number; it describes the specific unbounded behavior of the output.
The $\frac{c}{0}$ vs $\frac{0}{0}$ Diagnostic: Direct substitution yielding $\frac{c}{0}$ ( $c \neq 0$ ) establishes an infinite discontinuity and guarantees a vertical asymptote at $x = a$ . Conversely, $\frac{0}{0}$ represents an indeterminate form that may resolve to a finite hole or a vertical asymptote after factoring.
Signed Zero Directional Analysis: Determine the sign of the non-zero numerator and evaluate whether the denominator approaches zero through positive values ( $0^+$ ) or negative values ( $0^-$ ) to establish whether the limit diverges to $+\infty$ or $-\infty$ .

Key Equations & Structural Rules

Vertical Asymptote Definition: The line $x = a$ is a vertical asymptote if $\lim_{x \to a^-} f(x) = \pm\infty$ or $\lim_{x \to a^+} f(x) = \pm\infty$ .
Signed Zero Quotient Rules:
- $\frac{\text{Positive Constant}}{0^+} \to +\infty$
- $\frac{\text{Positive Constant}}{0^-} \to -\infty$
- $\frac{\text{Negative Constant}}{0^+} \to -\infty$
- $\frac{\text{Negative Constant}}{0^-} \to +\infty$

Worked Example A: Graphical Limit and Asymptote Identification

Consider the graph of a piecewise rational function $f(x)$ exhibiting asymptotes and boundary transitions:

As $x \to -4^-$ , the curve shoots upwards without bound.
As $x \to -4^+$ , the curve approaches an open circle at $(-4, 6)$ .
As $x \to 3^-$ , the curve shoots upwards without bound.
As $x \to 3^+$ , the curve shoots downwards without bound.
As $x \to \infty$ , the curve flattens along $y = -2$ .
As $x \to -\infty$ , the curve flattens along $y = 1$ .

(a) Evaluate $\lim_{x \to -4^-} f(x)$ , $\lim_{x \to -4^+} f(x)$ , $\lim_{x \to 3^-} f(x)$ , $\lim_{x \to 3^+} f(x)$ , $\lim_{x \to \infty} f(x)$ , and $\lim_{x \to -\infty} f(x)$ .
(b) Identify all vertical asymptotes.
(c) Identify all horizontal asymptotes.

💡 Reveal Worked Solution

Part (a) — Directional and End-Behavior Limits:
- $\lim_{x \to -4^-} f(x) = \infty$ (upward unbounded curve)
- $\lim_{x \to -4^+} f(x) = 6$ (approaches open hole coordinate)
- $\lim_{x \to 3^-} f(x) = \infty$ (upward unbounded curve)
- $\lim_{x \to 3^+} f(x) = -\infty$ (downward unbounded curve)
- $\lim_{x \to \infty} f(x) = -2$ (levels off along horizontal line)
- $\lim_{x \to -\infty} f(x) = 1$ (levels off along horizontal line)
Part (b) — Vertical Asymptotes:
- At $x = -4$ , the left-hand limit is $\infty$ , so $x = -4$ is a vertical asymptote.
- At $x = 3$ , both one-sided limits are infinite ( $\infty$ and $-\infty$ ), so $x = 3$ is a vertical asymptote.
Part (c) — Horizontal Asymptotes:
- As $x \to -\infty$ , $f(x) \to 1 \implies \mathbf{y = 1}$ .
- As $x \to \infty$ , $f(x) \to -2 \implies \mathbf{y = -2}$ .

Worked Example B: Analytical Asymptote and Discontinuity Classification

Using limits, find all vertical and horizontal asymptotes for:

f(x) = \frac{2x^2}{3x - x^2}

💡 Reveal Worked Solution

Factor and Identify Candidate Singularities: $f(x) = \frac{2x^2}{x(3 - x)}$ The denominator is zero at $x = 0$ and $x = 3$ .
Test Candidate $x = 0$ : Direct substitution gives $\frac{0}{0}$ . Canceling the common factor $x$ for $x \neq 0$ : $\lim_{x \to 0} \frac{2x}{3 - x} = \frac{2(0)}{3 - 0} = \frac{0}{3} = 0$ Because the limit is finite ( $0$ ), $x = 0$ is a removable discontinuity (hole), not a vertical asymptote.
Test Candidate $x = 3$ : Substituting into the simplified form $\frac{2x}{3 - x}$ :
$\lim_{x \to 3} \frac{2x}{3 - x} \to \frac{6}{0}$
Because this is $\frac{c}{0}$ ( $c \neq 0$ ), $x = 3$ is a vertical asymptote.
- Left-hand limit ( $x \to 3^-$ ): denominator $3 - x \to 0^+ \implies \lim_{x \to 3^-} f(x) = \frac{+6}{0^+} = +\infty$ .
- Right-hand limit ( $x \to 3^+$ ): denominator $3 - x \to 0^- \implies \lim_{x \to 3^+} f(x) = \frac{+6}{0^-} = -\infty$ .
Horizontal Asymptote Verification: Divide numerator and denominator by highest denominator power $x^2$ :
$\lim_{x \to \pm\infty} \frac{2x^2}{3x - x^2} = \lim_{x \to \pm\infty} \frac{2}{\frac{3}{x} - 1} = \frac{2}{0 - 1} = -2$
- Vertical Asymptote: $x = 3$
- Horizontal Asymptote: $y = -2$

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, what is the fundamental mathematical distinction between direct substitution yielding $\frac{c}{0}$ ( $c \neq 0$ ) versus yielding $\frac{0}{0}$ ?
(b) In Worked Example B, why is $x = 0$ classified as a removable discontinuity rather than a vertical asymptote, while $x = 3$ is classified as a vertical asymptote?
(c) Review the linked Engineering Applications section. How does a civil engineer use directional zero sign analysis to evaluate structural vibration amplitude in the Structural Resonance Threshold (Civil Engineering) model?

Practice Space Solve this problem on paper or tablet.

Review the $\frac{c}{0}$ vs $\frac{0}{0}$ rules in the Conceptual Framework, re-examine Steps 2 and 3 of Worked Example B, and read the introductory physical context of the Civil Engineering resonance problem.

Answer: * (a) $\frac{c}{0}$ ( $c \neq 0$ ) indicates an infinite discontinuity (vertical asymptote), whereas $\frac{0}{0}$ is an indeterminate form representing a removable hole (after factor cancellation) or requiring further algebraic simplification.

(b) At $x = 0$ , the zero-inducing factor cancels out giving a finite limit ( $\lim_{x \to 0} f(x) = 0$ ), whereas at $x = 3$ , the denominator remains zero while the numerator is non-zero ( $6/0$ ), producing unbounded infinite limits.
(c) As the excitation wind frequency approaches the natural frequency from below ( $\omega \to 8^-$ ), the denominator $(64 - \omega^2)$ approaches zero through positive values ( $0^+$ ), resulting in positive infinite amplitude escalation ( $\frac{+15}{0^+} = +\infty$ ).

Step-by-Step Execution

(a) A non-zero numerator over an infinitesimally small denominator ( $\frac{c}{0}$ ) grows without bound ( $\pm\infty$ ), defining a vertical asymptote. A $\frac{0}{0}$ form indicates that the numerator and denominator share a root, meaning the discontinuity is removable once the common factor is factored and canceled.
(b) Factoring $f(x) = \frac{2x^2}{x(3-x)} = \frac{2x}{3-x}$ (for $x \neq 0$ ) reveals that $\lim_{x \to 0} \frac{2x}{3-x} = 0$ . Since the limit exists and is finite, $x = 0$ is a removable hole. At $x = 3$ , substitution gives $\frac{6}{0}$ , which diverges to $\pm\infty$ , confirming a vertical asymptote.
(c) In structural engineering, approaching the resonance frequency $\omega_0 = 8\text{ rad/s}$ from below yields $\omega < 8 \implies \omega^2 < 64 \implies 64 - \omega^2 > 0$ . The signed zero approach gives $\frac{15}{0^+} = +\infty$ , proving vibration amplitude grows infinitely without damping.

Practice Problem 2

Evaluate the directional limit analytically:

\lim_{x \to 2^{+}} \frac{-5}{x-2}

Practice Space Solve this problem on paper or tablet.

Evaluate the sign of the numerator ( $-5$ ) and determine whether the denominator approaches zero through positive numbers ( $0^+$ ) or negative numbers ( $0^-$ ) as $x \to 2^+$ .

Answer: $-\infty$

Step-by-Step Execution

Numerator: $-5$ (negative constant).
Denominator: As $x \to 2^+$ , $x > 2$ , so $x - 2 > 0$ (approaches $0^+$ ).
Sign division: $\lim_{x \to 2^+} \frac{-5}{x-2} = \frac{-5}{0^+} = -\infty$ .

Practice Problem 3

Evaluate the directional limit analytically:

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

Practice Space Solve this problem on paper or tablet.

Evaluate the sign of the numerator ( $+4$ ) and determine whether the denominator approaches zero through positive numbers ( $0^+$ ) or negative numbers ( $0^-$ ) as $x \to 3^+$ .

Answer: $-\infty$

Step-by-Step Execution

Numerator: $+4$ (positive constant).
Denominator: As $x \to 3^+$ , $x > 3$ , so $3 - x < 0$ (approaches $0^-$ ).
Sign division: $\lim_{x \to 3^+} \frac{4}{3-x} = \frac{+4}{0^-} = -\infty$ .

Practice Problem 4

Evaluate the directional limit analytically:

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

Practice Space Solve this problem on paper or tablet.

Evaluate the sign of the numerator $3x$ at $x = -4$ and check whether the denominator approaches zero from the right ( $0^+$ ) or left ( $0^-$ ).

Answer: $-\infty$

Step-by-Step Execution

Numerator: As $x \to -4^+$ , $3x \to 3(-4) = -12 < 0$ .
Denominator: As $x \to -4^+$ , $x > -4$ , so $x + 4 > 0$ (approaches $0^+$ ).
Sign division: $\lim_{x \to -4^+} \frac{3x}{x+4} = \frac{-12}{0^+} = -\infty$ .

Practice Problem 5

Evaluate the limit analytically:

\lim_{x \to 0^{+}} \frac{1}{x+5}

Practice Space Solve this problem on paper or tablet.

Test whether direct substitution yields a non-zero denominator (a determinate form).

Answer: $\frac{1}{5}$

Step-by-Step Execution

Denominator: As $x \to 0^+$ , $x + 5 \to 5 \neq 0$ .
Direct substitution is determinate: $\lim_{x \to 0^+} \frac{1}{x+5} = \frac{1}{0+5} = \frac{1}{5}$ .

Practice Problem 6

Evaluate the limit analytically:

\lim_{x \to -5^{+}} \frac{(x+5)^{2}}{2x^{2}+10x}

Practice Space Solve this problem on paper or tablet.

Direct substitution gives $\frac{0}{0}$ . Factor the common term out of the denominator to cancel $(x + 5)$ before re-evaluating.

Answer: $0$

Step-by-Step Execution

Direct substitution gives $\frac{0}{0}$ . Factor the denominator: $2x^2 + 10x = 2x(x + 5)$ .
Simplify for $x \neq -5$ : $\frac{(x+5)^2}{2x(x+5)} = \frac{x+5}{2x}$
Re-evaluate the limit by direct substitution: $\lim_{x \to -5^+} \frac{x+5}{2x} = \frac{-5 + 5}{2(-5)} = \frac{0}{-10} = 0$

Practice Problem 7

Evaluate the trigonometric limit analytically:

\lim_{\theta \to \pi^{-}} \csc(\theta)

Practice Space Solve this problem on paper or tablet.

Rewrite using the reciprocal identity $\csc(\theta) = \frac{1}{\sin(\theta)}$ . Determine whether $\sin(\theta)$ approaches zero from positive or negative values in Quadrant II as $\theta \to \pi^-$ .

Answer: $\infty$

Step-by-Step Execution

Rewrite using the reciprocal identity: $\csc(\theta) = \frac{1}{\sin(\theta)}$ .
As $\theta \to \pi^-$ , $\theta$ is in Quadrant II ( $\theta < \pi$ ), where $\sin(\theta) > 0$ .
As $\theta \to \pi$ , $\sin(\theta) \to 0^+$ .
Sign division: $\lim_{\theta \to \pi^-} \frac{1}{\sin(\theta)} = \frac{1}{0^+} = \infty$ .

Segment 2 of 8

📅 Regroup 1

Key Reflection: Why does $\frac{c}{0}$ ( $c \neq 0$ ) indicate a vertical asymptote, while $\frac{0}{0}$ requires further algebraic work before classifying the discontinuity?
Directional Sign Check: Emphasize that evaluating directional limits at vertical asymptotes never requires plugging in long decimal tables—identifying whether the denominator is $0^+$ or $0^-$ and checking the numerator sign gives the exact sign of infinity immediately.

Segment 3 of 8

Skill Block 2

Real-Life Engineering Applications

🧪

Chemical Engineering

Industrial Bioreactor Output Bottlenecks (Chemical Engineering)

In an automated biochemical fermentation facility, process engineers monitor the production rate $V([S])$ (in kilograms of synthetic protein per hour) inside a continuous stirred-tank bioreactor. The reaction velocity follows a Michaelis-Menten saturation profile modeled as a function of the input substrate nutrient concentration $[S]$ (in millimoles per liter):

V([S]) = \frac{450[S]}{18 + [S]}

To optimize facility capital investment and feedstock supply logistics, the plant management team needs to determine the theoretical maximum production bottleneck ( $V_{\max}$ ) when the reactor is supplied with an unbounded excess of substrate ( $[S] \to \infty$ ).

Identify the highest powers of $[S]$ in the numerator and denominator to determine the growth dominance regime.
Evaluate the limit analytically: $\lim_{[S] \to \infty} \frac{450[S]}{18 + [S]}$
State the horizontal asymptote equation and its practical meaning for biochemical throughput.

💡 Show Worked Derivation

Part 1: Degree Comparison & Growth Dominance

The degree of $[S]$ in the numerator is $1$ (leading term $450[S]$ ), and the degree of $[S]$ in the denominator is $1$ (leading term $[S]$ ). Because the powers are equal ( $n = m = 1$ ), this rational function is in the balanced degrees regime, meaning the limit at infinity is dictated by the ratio of the leading coefficients.

Part 2: Analytical Limit Calculation

Divide numerator and denominator by the highest power of $[S]$ in the denominator, which is $[S]^1$ :

\lim_{[S] \to \infty} \frac{450[S]}{18 + [S]} = \lim_{[S] \to \infty} \frac{\frac{450[S]}{[S]}}{\frac{18}{[S]} + \frac{[S]}{[S]}} = \lim_{[S] \to \infty} \frac{450}{\frac{18}{[S]} + 1}

As $[S] \to \infty$ , the term $\frac{18}{[S]} \to 0$ :

\frac{450}{0 + 1} = 450\text{ kg/h}

Part 3: Horizontal Asymptote & Biochemical Interpretation

Horizontal Asymptote: $y = 450$
Physical Interpretation: The bioreactor has a strict upper throughput capacity of $V_{\max} = 450\text{ kg/h}$ . Regardless of how much additional substrate is pumped into the vessel, the enzymatic binding sites become fully saturated, making further feedstock expenditure economically ineffective beyond the asymptotic knee.

⚙️

General Engineering

Server Load Response Latency (Computer Science)

In a distributed cloud microservice architecture, software performance engineers evaluate the average network response latency $T(N)$ (in milliseconds) as a function of the concurrent active compute node count $N$ :

T(N) = \frac{12N^2 + 45N}{3N^2 + 8}

As the cluster auto-scales dynamically during peak traffic, the number of worker nodes becomes very large ( $N \to \infty$ ). Systems architects must determine the asymptotic latency ceiling to ensure the application complies with Service Level Agreements (SLAs).

Apply Growth Dominance filtering to identify the dominating terms in $T(N)$ .
Evaluate the limit analytically: $\lim_{N \to \infty} \frac{12N^2 + 45N}{3N^2 + 8}$
State the horizontal asymptote and explain how this value dictates cloud scaling limits.

💡 Show Worked Derivation

Part 1: Dominance Filtering

For large $N$ , the quadratic terms $N^2$ grow much faster than linear terms $N$ or constants.

Dominant numerator term: $12N^2$
Dominant denominator term: $3N^2$

Part 2: Analytical Limit Calculation

Divide every term by the highest denominator power $N^2$ :

\lim_{N \to \infty} \frac{\frac{12N^2}{N^2} + \frac{45N}{N^2}}{\frac{3N^2}{N^2} + \frac{8}{N^2}} = \lim_{N \to \infty} \frac{12 + \frac{45}{N}}{3 + \frac{8}{N^2}}

As $N \to \infty$ , the reciprocal terms approach zero:

\frac{12 + 0}{3 + 0} = \frac{12}{3} = 4\text{ ms}

Part 3: Horizontal Asymptote & Systems Architecture Conclusion

Horizontal Asymptote: $y = 4$
System SLA Impact: As the cluster scales indefinitely, the average request latency levels off at a deterministic floor of $4\text{ ms}$ . This informs cloud engineers that infrastructure latency will not degrade beyond $4\text{ ms}$ under heavy node concurrency, satisfying strict real-time user experience guarantees.

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

Limits at Infinity & Horizontal Asymptotes (Growth Dominance & Degree Taxonomy)

General Overview
Skill Block 2 focuses on evaluating the global end behavior of functions as $x \to \pm\infty$ , establishing the concept of Growth Dominance and mastering the rational degree taxonomy for horizontal asymptotes.

Conceptual Framework

Global End Behavior: Limits at infinity determine whether a function levels off along a horizontal line ( $y = L$ ) or diverges toward unbounded positive or negative values.
The Reciprocal Power Foundation: For any power $p > 0$ , $\lim_{x \to \pm\infty} \frac{c}{x^p} = 0$ . This fundamental truth allows us to divide every term by the highest power of the denominator.
Rational Degree Comparison Taxonomy:
- Bottom-Heavy ( $n < m$ ): Denominator degree exceeds numerator degree $\implies \lim_{x \to \pm\infty} f(x) = 0 \implies \text{HA: } y = 0$ .
- Balanced ( $n = m$ ): Numerator and denominator degrees are equal $\implies \lim_{x \to \pm\infty} f(x) = \frac{a_n}{b_m} \implies \text{HA: } y = \frac{a_n}{b_m}$ .
- Top-Heavy ( $n > m$ ): Numerator degree exceeds denominator degree $\implies \lim_{x \to \pm\infty} f(x) = \pm\infty \implies \text{No HA}$ .

Key Equations & Structural Rules

Limit at Infinity: $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ .
Horizontal Asymptote: The line $y = L$ is a horizontal asymptote if $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ .
Polynomial End-Behavior Dominance: $\lim_{x \to \pm\infty} (a_n x^n + a_{n-1} x^{n-1} + \dots + a_0) = \lim_{x \to \pm\infty} a_n x^n$ .

Worked Example A: Balanced Degree Rational Limit and Horizontal Asymptote

Evaluate the limit and state the equation of any horizontal asymptote:

\lim_{x \to \infty} \frac{-5x^{3}-2x+3}{3x^{3}+3x^{2}-5x}

💡 Reveal Worked Solution

Identify Denominator Dominant Power: The highest power of $x$ in the denominator is $x^3$ .
Divide Numerator and Denominator by $x^3$ : $\lim_{x \to \infty} \frac{\frac{-5x^3}{x^3} - \frac{2x}{x^3} + \frac{3}{x^3}}{\frac{3x^3}{x^3} + \frac{3x^2}{x^3} - \frac{5x}{x^3}} = \lim_{x \to \infty} \frac{-5 - \frac{2}{x^2} + \frac{3}{x^3}}{3 + \frac{3}{x} - \frac{5}{x^2}}$
Apply the Reciprocal Limit Rule ( $\lim_{x \to \infty} \frac{c}{x^p} = 0$ ): $\frac{-5 - 0 + 0}{3 + 0 - 0} = -\frac{5}{3}$
State Horizontal Asymptote:
- Limit: $-\frac{5}{3}$
- Horizontal Asymptote: $y = -\frac{5}{3}$

Worked Example B: Fractional and Radical Power Growth Dominance

Evaluate the limit and state any horizontal asymptote:

\lim_{x \to \infty} \frac{\sqrt[3]{x}+4}{1-5\sqrt{x}}

💡 Reveal Worked Solution

Convert Radicals to Fractional Exponents: $\sqrt[3]{x} = x^{1/3} \quad \text{and} \quad \sqrt{x} = x^{1/2}$ The highest power of $x$ in the denominator is $x^{1/2}$ .
Divide Numerator and Denominator by $x^{1/2}$ : $\lim_{x \to \infty} \frac{\frac{x^{1/3}}{x^{1/2}} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - \frac{5x^{1/2}}{x^{1/2}}} = \lim_{x \to \infty} \frac{x^{1/3 - 1/2} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - 5} = \lim_{x \to \infty} \frac{\frac{1}{x^{1/6}} + \frac{4}{x^{1/2}}}{\frac{1}{x^{1/2}} - 5}$
Evaluate Limits of Reciprocal Terms: Since $1/6 > 0$ and $1/2 > 0$ , both terms in the numerator vanish to $0$ : $\frac{0 + 0}{0 - 5} = \frac{0}{-5} = 0$
State Horizontal Asymptote:
- Limit: $0$
- Horizontal Asymptote: $y = 0$

Practice Problem 8: Concept & Reading Check

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

(a) In the Conceptual Framework, what are the three outcomes in the rational degree comparison taxonomy, and what does each outcome imply for horizontal asymptotes?
(b) In Worked Example B, how were the radical expressions rewritten and compared to determine which power of $x$ dominates in the denominator?
(c) Review the linked Engineering Applications section. How does growth dominance explain the maximum production speed ceiling $V_{\max}$ in the Industrial Bioreactor Output Bottlenecks (Chemical Engineering) model?

Practice Space Solve this problem on paper or tablet.

Review the Rational Degree Comparison Taxonomy table, re-read Step 1 of Worked Example B, and inspect Part 1 of the Chemical Engineering bioreactor application solution.

Answer: * (a) Bottom-Heavy ( $n < m \implies \text{limit is } 0, \text{HA: } y = 0$ ), Balanced ( $n = m \implies \text{limit is } a_n/b_m, \text{HA: } y = a_n/b_m$ ), Top-Heavy ( $n > m \implies \text{limit diverges to } \pm\infty, \text{No HA}$ ).

(b) Radicals were converted to fractional exponents ( $\sqrt[3]{x} = x^{1/3}$ and $\sqrt{x} = x^{1/2}$ ). Comparing powers showed the denominator has higher degree ( $1/2 > 1/3$ ), proving the fraction is bottom-heavy and vanishes to $0$ .
(c) As substrate concentration $[S] \to \infty$ , the degree of the numerator matches the degree of the denominator ( $n = m = 1$ ), meaning output velocity is dictated by the ratio of leading coefficients: $\frac{450[S]}{1[S]} = 450\text{ kg/h}$ .

Step-by-Step Execution

(a) When the denominator degree is strictly larger ( $n < m$ ), the fraction is bottom-heavy and shrinks to $0$ ( $y = 0$ ). When degrees are equal ( $n = m$ ), growth rates are balanced and the limit is the ratio of leading coefficients ( $y = a_n/b_m$ ). When the numerator degree is strictly larger ( $n > m$ ), the top grows faster and diverges to $\pm\infty$ (no horizontal asymptote).
(b) By rewriting radicals as powers of $x$ ( $x^{1/3}$ vs $x^{1/2}$ ), the student identifies $x^{1/2}$ as the highest power in the denominator. Dividing all terms by $x^{1/2}$ results in positive powers in the denominator of each numerator term ( $x^{1/2 - 1/3} = x^{1/6}$ ), driving the overall limit to $0$ .
(c) In enzyme saturation kinetics, the model $V([S]) = \frac{450[S]}{18 + [S]}$ has balanced degrees in $[S]$ . For very large substrate additions, lower-order constants ( $18$ ) become negligible compared to $[S]$ , producing a horizontal asymptote at $V_{\max} = 450\text{ kg/h}$ .

Practice Problem 9

Evaluate the polynomial limit at negative infinity analytically:

\lim_{x \to -\infty} (7x^{7}-5x^{3}+4x-1)

Practice Space Solve this problem on paper or tablet.

Extract the highest-power leading term ( $7x^7$ ) and determine the sign when raising a negative base to an odd exponent.

Answer: $-\infty$

Step-by-Step Execution

The leading term is $7x^7$ .
As $x \to -\infty$ , $x^7 \to (-\infty)^7 = -\infty$ (an odd power preserves the negative sign).
Multiplying by $+7$ : $\lim_{x \to -\infty} 7x^7 = -\infty$ .

Practice Problem 10

Evaluate the polynomial limit at negative infinity analytically:

\lim_{x \to -\infty} (1-2x+5x^{2}-17x^{3}-4x^{7})

Practice Space Solve this problem on paper or tablet.

Identify the highest-power leading term ( $-4x^7$ ) and evaluate its behavior as $x \to -\infty$ .

Answer: $\infty$

Step-by-Step Execution

The leading term (highest power) is $-4x^7$ .
As $x \to -\infty$ , $x^7 \to -\infty$ .
Multiplying by negative coefficient $-4$ : $\lim_{x \to -\infty} (-4x^7) = (-4)(-\infty) = \infty$ .

Practice Problem 11

Evaluate the limit and state any horizontal asymptote (if none, state that):

\lim_{x \to \infty} \frac{4x^{5}+6x^{2}-2}{5-4x+2x^{3}}

Practice Space Solve this problem on paper or tablet.

Compare the degree of the numerator ( $5$ ) to the degree of the denominator ( $3$ ) and divide by $x^3$ .

Answer: * Limit: $\infty$

Horizontal Asymptote: None

Step-by-Step Execution

Divide by Highest Denominator Power ( $x^3$ ): $\lim_{x \to \infty} \frac{\frac{4x^5}{x^3} + \frac{6x^2}{x^3} - \frac{2}{x^3}}{\frac{5}{x^3} - \frac{4x}{x^3} + \frac{2x^3}{x^3}} = \lim_{x \to \infty} \frac{4x^2 + \frac{6}{x} - \frac{2}{x^3}}{\frac{5}{x^3} - \frac{4x}{x^2} + 2}$
Evaluate Component Limits:
- Denominator approaches $0 - 0 + 2 = 2$ .
- Numerator is dominated by $4x^2 \to \infty$ . $\frac{\infty}{2} = \infty$
Conclusion:
- Limit: $\infty$
- Horizontal Asymptote: None (function is top-heavy and grows without bound).

Practice Problem 12

Evaluate the rational limit and state the horizontal asymptote:

\lim_{x \to \infty} \frac{8x^4 - 3x^2 + 1}{2x^4 + 7x - 5}

Practice Space Solve this problem on paper or tablet.

Observe that the degrees are balanced ( $4$ and $4$ ). Divide through by $x^4$ and evaluate the ratio of leading coefficients.

Answer: Limit: $4$ ; Horizontal Asymptote: $y = 4$

Step-by-Step Execution

Both numerator and denominator have degree $4$ (balanced).
Divide by $x^4$ : $\lim_{x \to \infty} \frac{8 - \frac{3}{x^2} + \frac{1}{x^4}}{2 + \frac{7}{x^3} - \frac{5}{x^4}} = \frac{8 - 0 + 0}{2 + 0 - 0} = \frac{8}{2} = 4$
Horizontal Asymptote: $y = 4$ .

Practice Problem 13

Evaluate the rational limit and state the horizontal asymptote:

\lim_{x \to \infty} \frac{14x + 9}{3x^2 - 5x + 2}

Practice Space Solve this problem on paper or tablet.

Observe that the denominator degree ( $2$ ) exceeds the numerator degree ( $1$ ). Divide through by the highest denominator power $x^2$ .

Answer: Limit: $0$ ; Horizontal Asymptote: $y = 0$

Step-by-Step Execution

Numerator degree is $1$ , denominator degree is $2$ (bottom-heavy).
Divide by $x^2$ : $\lim_{x \to \infty} \frac{\frac{14}{x} + \frac{9}{x^2}}{3 - \frac{5}{x} + \frac{2}{x^2}} = \frac{0 + 0}{3 - 0 + 0} = \frac{0}{3} = 0$
Horizontal Asymptote: $y = 0$ .

Segment 4 of 8

📅 Regroup 2

Key Reflection: When evaluating limits at infinity, why does dividing by the highest power of the denominator work universally, even for top-heavy and bottom-heavy functions?
Degree Taxonomy Check: Have students practice quickly identifying whether a rational function has a horizontal asymptote of $y = 0$ , $y = a_n/b_m$ , or no HA simply by scanning the leading polynomial terms.

Segment 5 of 8

Skill Block 3

Real-Life Engineering Applications

🚀

Aerospace Engineering

entry Capsule (Aerospace Engineering)

An aerospace engineering team modeling the atmospheric re-entry descent of an unmanned scientific capsule determines that its downward velocity $v(t)$ (in meters per second) as a function of descent time $t$ (in seconds) is governed by the rational exponential model:

v(t) = \frac{76e^{0.4t} - 12}{2e^{0.4t} + 5}

To verify that the drogue parachutes and main deceleration canopies can safely deploy without exceeding maximum dynamic pressure and tensile tear limits, the recovery team must determine the capsule's asymptotic terminal velocity ( $v_{\text{term}} = \lim_{t \to \infty} v(t)$ ).

Identify the dominant growth terms as $t \to \infty$ .
Evaluate the limit of the capsule's velocity analytically as $t \to \infty$ : $\lim_{t \to \infty} \frac{76e^{0.4t} - 12}{2e^{0.4t} + 5}$
State the horizontal asymptote equation and interpret its value for parachute structural design.

💡 Show Worked Derivation

Part 1: Exponential Growth Dominance

As $t \to \infty$ , the exponential terms $e^{0.4t}$ grow exponentially without bound, rendering the additive constants $-12$ and $+5$ completely negligible.

Part 2: Analytical Limit Calculation

Divide both numerator and denominator by the dominant exponential term in the denominator, $e^{0.4t}$ :

\lim_{t \to \infty} \frac{76e^{0.4t} - 12}{2e^{0.4t} + 5} = \lim_{t \to \infty} \frac{\frac{76e^{0.4t}}{e^{0.4t}} - \frac{12}{e^{0.4t}}}{\frac{2e^{0.4t}}{e^{0.4t}} + \frac{5}{e^{0.4t}}} = \lim_{t \to \infty} \frac{76 - 12e^{-0.4t}}{2 + 5e^{-0.4t}}

Since $\lim_{t \to \infty} e^{-0.4t} = 0$ :

\frac{76 - 0}{2 + 0} = \frac{76}{2} = 38\text{ m/s}

Part 3: Horizontal Asymptote & Structural Conclusion

Horizontal Asymptote: $y = 38$
Aerospace Engineering Significance: The capsule's descent speed stabilizes at an absolute terminal velocity of $38\text{ m/s}$ ( $\approx 136.8\text{ km/h}$ ). The parachute deployment actuators and Kevlar tether lines must be rated to withstand the aerodynamic drag forces associated with this maximum velocity threshold.

⚙️

General Engineering

Autonomous Vehicle Radar Distance Margin (Computer Engineering)

In an autonomous vehicle driver-assistance system, a computer engineering team develops a dual-frequency frequency-modulated continuous-wave (FMCW) radar algorithm to estimate target tracking separation distance. The spatial calibration offset $\Delta d(x)$ (in meters) between direct reflection and synthetic aperture reconstruction as longitudinal range $x$ (in meters) expands is modeled by:

\Delta d(x) = \sqrt{x^2 + 8x} - x

As the vehicle tracks obstacles at high speeds over long highway distances ( $x \to \infty$ ), the team needs to know if the spatial offset diverges, vanishes, or settles to a constant calibration bias.

Explain why direct substitution of $x \to \infty$ results in an indeterminate form.
Evaluate the limit of the distance offset analytically using radical conjugate rationalization: $\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right)$
State the horizontal asymptote and explain how this constant offset is programmed into the vehicle's collision-avoidance radar firmware.

💡 Show Worked Derivation

Part 1: Indeterminate Form Classification

Direct substitution yields:

\sqrt{\infty^2 + 8(\infty)} - \infty = \infty - \infty

Because both competing terms grow without bound, this is an indeterminate form (

\infty - \infty

). The subtraction cannot be evaluated directly without algebraic manipulation.

Part 2: Analytical Limit Calculation via Conjugate Multiplication

Multiply and divide by the radical conjugate $(\sqrt{x^2 + 8x} + x)$ :

\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right) = \lim_{x \to \infty} \frac{(\sqrt{x^2 + 8x} - x)(\sqrt{x^2 + 8x} + x)}{\sqrt{x^2 + 8x} + x}

= \lim_{x \to \infty} \frac{(x^2 + 8x) - x^2}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{8x}{\sqrt{x^2 + 8x} + x}

Divide numerator and denominator by $x$ (with $x = \sqrt{x^2}$ since $x > 0$ ):

= \lim_{x \to \infty} \frac{8}{\sqrt{\frac{x^2 + 8x}{x^2}} + 1} = \lim_{x \to \infty} \frac{8}{\sqrt{1 + \frac{8}{x}} + 1}

As $x \to \infty$ , $\frac{8}{x} \to 0$ :

= \frac{8}{\sqrt{1 + 0} + 1} = \frac{8}{2} = 4\text{ m}

Part 3: Horizontal Asymptote & Firmware Calibration

Horizontal Asymptote: $y = 4$
Computer Engineering Application: At long ranges, the distance measurement converges to a steady systematic lead error of exactly $4\text{ meters}$ . Radar firmware engineers apply a static calibration correction offset of $-4\text{ m}$ in the tracking algorithm to ensure obstacle distance estimates remain accurate during highway driving.

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

Advanced End Behavior: Conjugates at Infinity, Negative Infinity with Radicals, & Exponential Models

General Overview
Skill Block 3 tackles advanced and subtle limit behaviors at infinity: resolving indeterminate $\infty - \infty$ radical differences using conjugates, applying the negative sign rule for $\sqrt{x^2} = |x| = -x$ as $x \to -\infty$ , and analyzing bounded exponential saturation models.

Conceptual Framework

Indeterminate Forms at Infinity ( $\infty - \infty$ ): Subtracting two quantities that both grow to infinity does not equal zero. We must multiply and divide by the radical conjugate to transform the expression into a manageable quotient.
The Absolute Value Trap as $x \to -\infty$ : When $x < 0$ , factoring $x^2$ out of a square root produces $\sqrt{x^2} = |x| = -x$ . Dividing inside a radical by $x^2$ when $x < 0$ requires placing a negative sign outside the radical: $\frac{\sqrt{A}}{x} = -\sqrt{\frac{A}{x^2}} \quad (\text{for } x < 0)$
Exponential End Behavior: The exponential term $e^{kx}$ ( $k > 0$ ) grows without bound as $x \to \infty$ and vanishes to $0$ as $x \to -\infty$ , while $e^{-kx}$ vanishes to $0$ as $x \to \infty$ .

Key Equations & Structural Rules

Radical Conjugate Identity: $\sqrt{A} - B = \frac{(\sqrt{A} - B)(\sqrt{A} + B)}{\sqrt{A} + B} = \frac{A - B^2}{\sqrt{A} + B}$
Negative Infinity Radical Sign Identity: $\sqrt{x^2} = -x$ for all $x < 0$ .
Exponential Limits at Infinity:
- $\lim_{x \to \infty} e^{-kx} = 0 \quad (k > 0)$
- $\lim_{x \to -\infty} e^{kx} = 0 \quad (k > 0)$

Worked Example A: Indeterminate Radical Form at Infinity ( $\infty - \infty$ )

Evaluate the limit:

\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right)

💡 Reveal Worked Solution

Recognize Indeterminate Form: Direct substitution gives $\sqrt{\infty^2 + \infty} - \infty = \infty - \infty$ (indeterminate).
Multiply and Divide by the Conjugate $(\sqrt{x^2 + 8x} + x)$ : $\lim_{x \to \infty} \frac{(\sqrt{x^2 + 8x} - x)(\sqrt{x^2 + 8x} + x)}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{(x^2 + 8x) - x^2}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{8x}{\sqrt{x^2 + 8x} + x}$
Divide Numerator and Denominator by $x$ ( $x = \sqrt{x^2}$ for $x > 0$ ): $\lim_{x \to \infty} \frac{\frac{8x}{x}}{\frac{\sqrt{x^2 + 8x}}{\sqrt{x^2}} + \frac{x}{x}} = \lim_{x \to \infty} \frac{8}{\sqrt{1 + \frac{8}{x}} + 1}$
Evaluate the Limit: $\frac{8}{\sqrt{1 + 0} + 1} = \frac{8}{1 + 1} = \frac{8}{2} = 4$

Worked Example B: Radical Limit at Negative Infinity ( $x \to -\infty$ )

Evaluate the limit and state the horizontal asymptote as $x \to -\infty$ :

\lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{2x - 5}

💡 Reveal Worked Solution

Identify the Sign of $x$ and Denominator Dominance: As $x \to -\infty$ , $x$ is negative ( $x < 0$ ). The dominant denominator power is $x^1$ .
Apply the Negative Radical Rule ( $\sqrt{x^2} = -x \implies x = -\sqrt{x^2}$ ): Dividing the numerator by $x$ (for $x < 0$ ): $\frac{\sqrt{9x^2 + 4}}{x} = \frac{\sqrt{9x^2 + 4}}{-\sqrt{x^2}} = -\sqrt{\frac{9x^2 + 4}{x^2}} = -\sqrt{9 + \frac{4}{x^2}}$
Divide Denominator by $x$ : $\frac{2x - 5}{x} = 2 - \frac{5}{x}$
Evaluate the Limit:
$\lim_{x \to -\infty} \frac{-\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{5}{x}} = \frac{-\sqrt{9 + 0}}{2 - 0} = -\frac{3}{2}$
- Limit: $-\frac{3}{2}$
- Horizontal Asymptote ( $x \to -\infty$ ): $y = -\frac{3}{2}$

Practice Problem 14: Concept & Reading Check

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

(a) Under the Conceptual Framework, why is the expression $\infty - \infty$ classified as an indeterminate form rather than evaluating to $0$ ? How does the conjugate method resolve this ambiguity?
(b) In Worked Example B, why did dividing $\sqrt{9x^2 + 4}$ by $x$ introduce an explicit negative sign ( $-\sqrt{9 + 4/x^2}$ )?
(c) Review the linked Engineering Applications section. How do aerospace engineers utilize limits at infinity of exponential models to determine the terminal velocity in the Terminal Velocity of a Re-entry Capsule (Aerospace Engineering) model?

Practice Space Solve this problem on paper or tablet.

Review the definition of $\infty - \infty$ in the Conceptual Framework, check Step 2 of Worked Example B regarding the $\sqrt{x^2} = |x|$ identity, and read Part 2 of the Aerospace Engineering terminal velocity solution.

Answer: * (a) $\infty - \infty$ is indeterminate because both terms grow without bound at independent rates; the difference depends entirely on their relative growth rate. Multiplying by the conjugate creates a rational quotient where leading powers can be compared directly.

(b) Because $x \to -\infty$ , $x$ is strictly negative ( $x < 0$ ). For negative numbers, $x = -\sqrt{x^2}$ . Bringing $1/x$ into the square root leaves a negative sign outside the radical.
(c) As time $t \to \infty$ , the exponential term $e^{0.4t}$ dominates constant offsets. Dividing through by $e^{0.4t}$ reveals that negative exponential terms ( $e^{-0.4t}$ ) vanish to $0$ , leaving the steady-state ratio of coefficients: $76/2 = 38\text{ m/s}$ .

Step-by-Step Execution

(a) In calculus, infinity is not a fixed real number; subtracting two unbounded expressions leaves open whether one grows faster, both grow at the same rate, or they differ by a constant. Multiplying by $\frac{\sqrt{A}+B}{\sqrt{A}+B}$ eliminates the radical difference in the numerator, producing a standard ratio that can be evaluated via Growth Dominance.
(b) For any real number, $\sqrt{x^2} = |x|$ . When $x < 0$ , $|x| = -x$ , meaning $x = -\sqrt{x^2}$ . Therefore, $\frac{\sqrt{9x^2+4}}{x} = \frac{\sqrt{9x^2+4}}{-\sqrt{x^2}} = -\sqrt{\frac{9x^2+4}{x^2}} = -\sqrt{9 + \frac{4}{x^2}}$ .
(c) In the capsule model $v(t) = \frac{76e^{0.4t}-12}{2e^{0.4t}+5}$ , dividing by $e^{0.4t}$ gives $\frac{76 - 12e^{-0.4t}}{2 + 5e^{-0.4t}}$ . As $t \to \infty$ , $e^{-0.4t} \to 0$ , giving the terminal velocity $v_{\text{term}} = \frac{76}{2} = 38\text{ m/s}$ .

Practice Problem 15

Evaluate the limit analytically using conjugate rationalization:

\lim_{x \to \infty} \left( \sqrt{4x^2 + 12x} - 2x \right)

Practice Space Solve this problem on paper or tablet.

Multiply and divide by the conjugate expression $(\sqrt{4x^2 + 12x} + 2x)$ and simplify the difference of squares in the numerator.

Answer:

3

Step-by-Step Execution

Multiply and Divide by the Conjugate: $\lim_{x \to \infty} \frac{(\sqrt{4x^2 + 12x} - 2x)(\sqrt{4x^2 + 12x} + 2x)}{\sqrt{4x^2 + 12x} + 2x} = \lim_{x \to \infty} \frac{(4x^2 + 12x) - 4x^2}{\sqrt{4x^2 + 12x} + 2x} = \lim_{x \to \infty} \frac{12x}{\sqrt{4x^2 + 12x} + 2x}$
Divide Numerator and Denominator by $x$ ( $x = \sqrt{x^2}$ for $x > 0$ ): $\lim_{x \to \infty} \frac{12}{\sqrt{\frac{4x^2 + 12x}{x^2}} + 2} = \lim_{x \to \infty} \frac{12}{\sqrt{4 + \frac{12}{x}} + 2}$
Evaluate the Limit: $= \frac{12}{\sqrt{4 + 0} + 2} = \frac{12}{2 + 2} = \frac{12}{4} = 3$

Practice Problem 16

Evaluate the function $f(x) = \frac{6x - 1}{\sqrt{4x^2 + 9}}$ as:

(a) $x \to \infty$
(b) $x \to -\infty$
(c) State all horizontal asymptotes of $f(x)$ .

Practice Space Solve this problem on paper or tablet.

For (a), use $x = \sqrt{x^2}$ since $x > 0$ . For (b), use $x = -\sqrt{x^2}$ since $x < 0$ .

Answer: * (a) $3$

(b) $-3$
(c) $y = 3$ and $y = -3$

Step-by-Step Execution

Part (a) — As $x \to \infty$ ( $x > 0 \implies x = \sqrt{x^2}$ ): $\lim_{x \to \infty} \frac{6 - \frac{1}{x}}{\sqrt{\frac{4x^2 + 9}{x^2}}} = \lim_{x \to \infty} \frac{6 - \frac{1}{x}}{\sqrt{4 + \frac{9}{x^2}}} = \frac{6 - 0}{\sqrt{4 + 0}} = \frac{6}{2} = 3$
Part (b) — As $x \to -\infty$ ( $x < 0 \implies x = -\sqrt{x^2}$ ): $\lim_{x \to -\infty} \frac{6 - \frac{1}{x}}{-\sqrt{\frac{4x^2 + 9}{x^2}}} = \lim_{x \to -\infty} \frac{6 - \frac{1}{x}}{-\sqrt{4 + \frac{9}{x^2}}} = \frac{6 - 0}{-\sqrt{4 + 0}} = \frac{6}{-2} = -3$
Part (c) — Horizontal Asymptotes:
- Right-side asymptote: $y = 3$
- Left-side asymptote: $y = -3$

Practice Problem 17

For the saturation model $g(x) = \frac{24}{3 + 5e^{-0.2x}}$ :

(a) Evaluate $\lim_{x \to \infty} g(x)$ .
(b) Evaluate $\lim_{x \to -\infty} g(x)$ .
(c) State all horizontal asymptotes of $g(x)$ .

Practice Space Solve this problem on paper or tablet.

Recall that as $x \to \infty$ , $e^{-0.2x} \to 0$ . As $x \to -\infty$ , $-0.2x \to +\infty$ , so $e^{-0.2x} \to \infty$ .

Answer: * (a) $8$

(b) $0$
(c) $y = 8$ and $y = 0$

Step-by-Step Execution

Part (a) — As $x \to \infty$ : Since $\lim_{x \to \infty} e^{-0.2x} = 0$ : $\lim_{x \to \infty} \frac{24}{3 + 5e^{-0.2x}} = \frac{24}{3 + 5(0)} = \frac{24}{3} = 8$
Part (b) — As $x \to -\infty$ : As $x \to -\infty$ , $-0.2x \to +\infty$ , which means $e^{-0.2x} \to \infty$ . $\lim_{x \to -\infty} \frac{24}{3 + 5e^{-0.2x}} \to \frac{24}{\infty} = 0$
Part (c) — Horizontal Asymptotes:
- As $x \to \infty$ : $y = 8$
- As $x \to -\infty$ : $y = 0$

Segment 6 of 8

📅 Regroup 3

Key Reflection: Why does the function $f(x) = \frac{6x-1}{\sqrt{4x^2+9}}$ possess two distinct horizontal asymptotes ( $y = 3$ and $y = -3$ ), whereas standard polynomial rational functions possess at most one?
Conjugate Rationalization Strategy: Emphasize that whenever students encounter $\infty - \infty$ with square roots, conjugate multiplication converts subtraction into addition in the denominator, eliminating ambiguity.

Segment 7 of 8

📅 Synthesis Wrap-up

Mastery Summary:
- Vertical Asymptotes: Occur at $x = a$ when $\lim_{x \to a^\pm} f(x) = \pm\infty$ ( $\frac{c}{0}$ form). Use directional signed zero analysis ( $0^+$ vs $0^-$ ).
- Horizontal Asymptotes: Occur at $y = L$ when $\lim_{x \to \pm\infty} f(x) = L$ . Filter using Growth Dominance (bottom-heavy $\to 0$ , balanced $\to a_n/b_m$ , top-heavy $\to \pm\infty$ ).
- Radicals & Exponentials: Use conjugates for $\infty - \infty$ , remember $\sqrt{x^2} = -x$ for $x < 0$ , and evaluate exponential decays to steady-state ceilings.
Knewton Alta Bridge: Completes the foundation for homework assignment Infinite Limits and Asymptotes and prepares students for Session 05: Special Limits and Continuity.

Segment 8 of 8

Infinite Limits and Limits at Infinity

📅 Micro-Lecture

Skill Block 1

Real-Life Engineering Applications

Structural Resonance Threshold (Civil Engineering)

Thermal Runaway in Electrical Discharges (Electrical Engineering)

Worked Example A: Graphical Limit and Asymptote Identification

Worked Example B: Analytical Asymptote and Discontinuity Classification

Practice Problem 1: Concept & Reading Check

Step-by-Step Execution

Practice Problem 2

Step-by-Step Execution

Practice Problem 3

Step-by-Step Execution

Practice Problem 4

Step-by-Step Execution

Practice Problem 5

Step-by-Step Execution

Practice Problem 6

Step-by-Step Execution

Practice Problem 7

Step-by-Step Execution

📅 Regroup 1

Skill Block 2

Real-Life Engineering Applications

Industrial Bioreactor Output Bottlenecks (Chemical Engineering)

Server Load Response Latency (Computer Science)

Worked Example A: Balanced Degree Rational Limit and Horizontal Asymptote

Worked Example B: Fractional and Radical Power Growth Dominance

Practice Problem 8: Concept & Reading Check

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

Practice Problem 12

Step-by-Step Execution

Practice Problem 13

Step-by-Step Execution

📅 Regroup 2

Skill Block 3

Real-Life Engineering Applications

entry Capsule (Aerospace Engineering)

Autonomous Vehicle Radar Distance Margin (Computer Engineering)

Worked Example A: Indeterminate Radical Form at Infinity (∞−∞\infty - \infty∞−∞)

Worked Example B: Radical Limit at Negative Infinity (x→−∞x \to -\inftyx→−∞)

Practice Problem 14: Concept & Reading Check

Step-by-Step Execution

Practice Problem 15

Step-by-Step Execution

Practice Problem 16

Step-by-Step Execution

Practice Problem 17

Step-by-Step Execution

📅 Regroup 3

📅 Synthesis Wrap-up

Worked Example A: Indeterminate Radical Form at Infinity ( $\infty - \infty$ )

Worked Example B: Radical Limit at Negative Infinity ( $x \to -\infty$ )