MATH-181 (Calculus I) Curriculum

Session 12 - L’Hopital’s Rule and Indeterminate Forms

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Diagnose and classify fractional indeterminate forms ( $\frac{0}{0}$ and $\frac{\pm\infty}{\pm\infty}$ ), execute the strict "STOP & CHECK" protocol to distinguish indeterminate forms from determinate limits, and apply L'Hôpital's Rule (including successive repeated derivatives) by differentiating numerators and denominators independently.
Resolve indeterminate products ( $0 \cdot \pm\infty$ ) and indeterminate differences ( $\infty - \infty$ ) by executing strategic algebraic inversions, trigonometric reciprocal transformations, and common denominator unifications to generate viable fractional quotients for L'Hôpital evaluation.
Execute the 4-step Logarithmic Transformation Algorithm ( $y = [f(x)]^{g(x)} \implies \ln y \implies \lim \ln y = L \implies \lim y = e^L$ ) to evaluate indeterminate power forms ( $1^\infty$ , $0^0$ , $\infty^0$ ), avoiding the halfway trap by correctly exponentiating the final limit.

Micro-Lecture

1. The Physics and Geometry of Indeterminate Quotients

Mathematical Foundation: Direct substitution into a limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ frequently produces expressions like $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ . These expressions are indeterminate forms—they indicate that the limiting ratio depends on the relative rates at which the numerator and denominator approach zero or infinity. Through local linear approximation (tangent line linearization), if $f(a) = 0$ and $g(a) = 0$ , then near $x = a$ : $\frac{f(x)}{g(x)} \approx \frac{f(a) + f'(a)(x - a)}{g(a) + g'(a)(x - a)} = \frac{0 + f'(a)(x - a)}{0 + g'(a)(x - a)} = \frac{f'(a)}{g'(a)}$ This establishes L'Hôpital's Rule: when direct substitution yields $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ , the limit of the quotient equals the limit of the quotient of their derivatives: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (see L'Hôpital's Rule).
Engineering Context: In high-speed telecommunications and radar signal processing, transmitted waveform pulses are governed by the sinc function $h(t) = \frac{\sin(\pi B t)}{\pi B t}$ . At the center of the pulse ( $t = 0$ ), direct evaluation gives $\frac{0}{0}$ . Applying L'Hôpital's Rule establishes that the normalized peak pulse amplitude is exactly $1$ , allowing DSP firmware to scale analog-to-digital converters and avoid run-time division-by-zero crashes.

2. The Strict "STOP & CHECK" Protocol vs. The Quotient Rule Confusion

Mathematical Foundation: Two major operational traps derail students on exams:
- The Determinate Trap: Applying L'Hôpital's Rule to a determinate form yields garbage. For example, $\lim_{x \to 0} \frac{x+2}{x+1} = \frac{2}{1} = 2$ . Blindly applying L'Hôpital gives $\lim_{x \to 0} \frac{1}{1} = 1$ , which is completely false! You must verify the presence of $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ before differentiating.
- The Quotient Rule Confusion: In L'Hôpital's Rule, you never use the Quotient Rule. You differentiate the numerator function $f(x)$ and denominator function $g(x)$ separately: $\frac{f'(x)}{g'(x)}$ , not $\frac{f'g - fg'}{g^2}$ .

3. The Hierarchy of Indeterminate Forms & Asymptotic Growth

Mathematical Foundation: Beyond quotients, calculus presents five other indeterminate forms: products ( $0 \cdot \pm\infty$ ), differences ( $\infty - \infty$ ), and powers ( $1^\infty, 0^0, \infty^0$ ) (see Indeterminate Forms). Furthermore, as $x \to \infty$ , functions adhere to a strict asymptotic dominance hierarchy: $\ln x \ll x^p \ll a^x \ll x!$ L'Hôpital's Rule rigorously demonstrates that any exponential function outgrows any polynomial, and any polynomial outgrows any logarithm (see Growth Rates of Functions).

Skill Block 1

Worked Example A: Indeterminate Quotient with Chain Rule

Evaluate the limit:

\lim_{x \to 0} \frac{\sin(4x)}{2x}

Worked Solution:

Name the Governing Formula: L'Hôpital's Rule for Indeterminate Quotients: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (This rule is defined in L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- Numerator: $\sin(4 \cdot 0) = \sin(0) = 0$ .
- Denominator: $2(0) = 0$ .
- The expression is an indeterminate quotient of form $\frac{0}{0}$ . L'Hôpital's Rule is applicable.
Differentiate Numerator and Denominator Independently:
- Differentiate the numerator using the Chain Rule: $\frac{d}{dx}[\sin(4x)] = \cos(4x) \cdot \frac{d}{dx}[4x] = 4\cos(4x)$
- Differentiate the denominator: $\frac{d}{dx}[2x] = 2$
Form the New Limit and Evaluate: $\lim_{x \to 0} \frac{\sin(4x)}{2x} \overset{\text{L'H}}{=} \lim_{x \to 0} \frac{4\cos(4x)}{2} = \frac{4\cos(0)}{2} = \frac{4(1)}{2} = 2$

Worked Example B: Repeated L'Hôpital on an Exponential Quotient

Evaluate the limit:

\lim_{x \to \infty} \frac{x^2}{e^{3x}}

Worked Solution:

Name the Governing Formula: Successive Applications of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{f''(x)}{g''(x)}$ (This rule is defined in L'Hôpital's Rule and Growth Rates of Functions.)
Execute the STOP & CHECK Protocol:
- Numerator: $\lim_{x \to \infty} x^2 = \infty$ .
- Denominator: $\lim_{x \to \infty} e^{3x} = \infty$ .
- The expression is an indeterminate quotient of form $\frac{\infty}{\infty}$ .
First Application of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{x^2}{e^{3x}} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{2x}{3e^{3x}}$
Re-check the Form:
- Numerator: $2(\infty) = \infty$ .
- Denominator: $3e^\infty = \infty$ .
- The form is still $\frac{\infty}{\infty}$ . L'Hôpital's Rule applies a second time.
Second Application of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{2x}{3e^{3x}} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{2}{9e^{3x}}$
Evaluate the Determinate Limit:
- As $x \to \infty$ , the numerator remains constant ( $2$ ) while the denominator diverges to $\infty$ : $\frac{2}{\infty} = 0$
- Engineering significance: This proves that the exponential growth in the denominator completely overwhelms the polynomial growth of the numerator.

Practice Problem 1: Concept & Reading Check

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

(a) In the Conceptual Framework, explain the "STOP & CHECK" protocol. What happens if an engineer applies L'Hôpital's Rule to $\lim_{x \to 0} \frac{x+2}{x+1}$ ?
(b) Deconstruct Worked Example B: Why is it necessary to apply L'Hôpital's Rule twice? Why would using the Quotient Rule $\frac{f'g - fg'}{g^2}$ to evaluate this limit be mathematically incorrect?
(c) Review the linked Engineering Applications (Digital Signal Sinc Pulse Filtering (Electrical & Computer Engineering)). Why does direct evaluation of the sinc pulse at $t = 0$ cause an exception in DSP microcontrollers, and how does L'Hôpital's Rule resolve the peak voltage?

Solution Key:

(a) The STOP & CHECK protocol enforces that L'Hôpital's Rule is an emergency diagnostic tool applicable strictly to indeterminate quotients ( $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ ). If direct substitution yields a determinate real number, that number is the exact value of the limit. Evaluating $\lim_{x \to 0} \frac{x+2}{x+1}$ by direct substitution gives $\frac{0+2}{0+1} = \frac{2}{1} = 2$ . If one erroneously applies L'Hôpital's Rule by differentiating numerator and denominator, one obtains $\lim_{x \to 0} \frac{1}{1} = 1$ , which is mathematically false and introduces a 50% calculation error into an engineering design.
(b) In Worked Example B, the first application of L'Hôpital yields $\lim_{x \to \infty} \frac{2x}{3e^{3x}}$ . Because the numerator still grows without bound ( $2x \to \infty$ ) and the denominator still grows without bound ( $3e^{3x} \to \infty$ ), the form remains $\frac{\infty}{\infty}$ , requiring a second application to reduce the polynomial numerator to a constant. The Quotient Rule is strictly the formula for the derivative of a combined quotient function ( $\frac{d}{dx}[\frac{f}{g}] = \frac{f'g - fg'}{g^2}$ ). L'Hôpital's Rule is not finding the derivative of the fraction; it is computing the ratio of instantaneous linear slopes ( $\frac{f'(x)}{g'(x)}$ ) to determine the limiting value.
(c) In embedded digital signal processors and FPGAs, evaluating $\frac{\sin(\pi B t)}{\pi B t}$ at the sampling clock instant $t = 0$ requires the ALU to execute division by zero, triggering an IEEE 754 NaN exception or an integer divide-by-zero hardware fault that crashes the receiver firmware. L'Hôpital's Rule evaluates the removable singularity by taking the ratio of derivatives ( $\frac{\pi B \cos(\pi B t)}{\pi B} \to 1$ ), proving that the true physical pulse amplitude at $t = 0$ is exactly $1.0\text{ V}$ (or scaled to $A_0 = 3.3\text{ V}$ ). This allows firmware engineers to safely insert a branchless conditional check or hardcode the zero-index FIR filter tap coefficient.

Show your steps and calculations here.

Practice Problem 2

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to 1} \frac{\sin(1 - x^3)}{x^2 - 1}

Solution Key:

Name the Governing Formula: L'Hôpital's Rule for Indeterminate Quotients: $\lim_{x \to a} \frac{f(x)}{g(x)} \overset{\text{L'H}}{=} \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (This rule is defined in L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- Numerator at $x = 1$ : $\sin(1 - 1^3) = \sin(0) = 0$ .
- Denominator at $x = 1$ : $1^2 - 1 = 0$ .
- Indeterminate form: $\frac{0}{0}$ . L'Hôpital's Rule applies.
Differentiate Numerator and Denominator Independently:
- Numerator via Chain Rule: $\frac{d}{dx}[\sin(1 - x^3)] = \cos(1 - x^3) \cdot \frac{d}{dx}[1 - x^3] = \cos(1 - x^3) \cdot (-3x^2) = -3x^2\cos(1 - x^3)$
- Denominator via Power Rule: $\frac{d}{dx}[x^2 - 1] = 2x$
Evaluate the Limit: $\lim_{x \to 1} \frac{\sin(1 - x^3)}{x^2 - 1} \overset{\text{L'H}}{=} \lim_{x \to 1} \frac{-3x^2\cos(1 - x^3)}{2x}$ Substitute $x = 1$ : $\frac{-3(1)^2 \cos(1 - 1)}{2(1)} = \frac{-3(1)\cos(0)}{2} = \frac{-3(1)(1)}{2} = -\frac{3}{2}$

Show your steps and calculations here.

Practice Problem 3

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to 1} \frac{x^2 - 16}{3 - 5\ln x}

Solution Key:

Name the Governing Formula: Direct Substitution and the STOP & CHECK Protocol: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)} \quad \text{if } g(a) \neq 0$ (This rule is defined in L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- Numerator at $x = 1$ : $1^2 - 16 = 1 - 16 = -15$ .
- Denominator at $x = 1$ : $3 - 5\ln(1) = 3 - 5(0) = 3$ .
Classify the Limit:
- The quotient is: $\frac{-15}{3} = -5$
- This form is determinate. L'Hôpital's Rule is NOT applicable.
Final Conclusion: $\lim_{x \to 1} \frac{x^2 - 16}{3 - 5\ln x} = -5$ (Warning: Applying L'Hôpital's Rule erroneously gives $\lim_{x \to 1}\frac{2x}{-5/x} = -\frac{2}{5}$ , which is incorrect.)

Show your steps and calculations here.

Practice Problem 4

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to \infty} \frac{e^{5x}}{x^2}

Solution Key:

Name the Governing Formula: Successive Applications of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{f(x)}{g(x)} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{f''(x)}{g''(x)}$ (This rule is defined in L'Hôpital's Rule and Growth Rates of Functions.)
Execute the STOP & CHECK Protocol:
- Numerator: $\lim_{x \to \infty} e^{5x} = \infty$ .
- Denominator: $\lim_{x \to \infty} x^2 = \infty$ .
- Indeterminate form: $\frac{\infty}{\infty}$ . L'Hôpital's Rule applies.
First Application of L'Hôpital's Rule: Differentiate numerator using Chain Rule and denominator using Power Rule: $\lim_{x \to \infty} \frac{e^{5x}}{x^2} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{5e^{5x}}{2x}$
Re-check Form:
- As $x \to \infty$ , $5e^{5x} \to \infty$ and $2x \to \infty$ . Form is still $\frac{\infty}{\infty}$ .
Second Application of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{5e^{5x}}{2x} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{25e^{5x}}{2}$
Evaluate the Limit: As $x \to \infty$ , $25e^{5x} \to \infty$ , while the denominator is the static constant $2$ : $\lim_{x \to \infty} \frac{25e^{5x}}{2} = \infty$ The limit diverges to $+\infty$ .

Show your steps and calculations here.

Regroup 1

The STOP & CHECK Discipline: Reinforce that students must visibly write down the substitution test on their worksheets before applying L'Hôpital's Rule. Emphasize that finding a determinate limit (like in Practice Problem 3) is a victory that saves time and prevents catastrophic calculation errors.
Independence of Derivatives: Remind students that L'Hôpital evaluates $\frac{f'(x)}{g'(x)}$ ; never write down the Quotient Rule $\frac{f'g - fg'}{g^2}$ when evaluating an indeterminate limit.
Algebraic Cleanup Before Next Differentiation: When repeated L'Hôpital is necessary, simplify constants and exponents after each derivative step before differentiating again.

Skill Block 2

Worked Example A: Indeterminate Product Inversion

Evaluate the limit:

\lim_{x \to 0^+} x^2 \ln(x)

Worked Solution:

Name the Governing Formula: Reciprocal Inversion for Indeterminate Products: $f(x) \cdot g(x) = \frac{g(x)}{[f(x)]^{-1}}$ (This strategy is defined in Indeterminate Forms and L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- As $x \to 0^+$ , $x^2 \to 0$ and $\ln(x) \to -\infty$ .
- Indeterminate form: $0 \cdot (-\infty)$ . L'Hôpital's Rule cannot be applied directly.
Restructure as a Fractional Quotient: Keep $\ln(x)$ in the numerator and invert $x^2$ into the denominator as $x^{-2}$ : $\lim_{x \to 0^+} x^2 \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{x^{-2}}$ Re-checking form: $\frac{-\infty}{\infty}$ . We have a green light for L'Hôpital's Rule.
Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{\ln(x)}{x^{-2}} \overset{\text{L'H}}{=} \lim_{x \to 0^+} \frac{\frac{1}{x}}{-2x^{-3}}$
Simplify Algebraically Before Taking Limit: $\frac{\frac{1}{x}}{-\frac{2}{x^3}} = \frac{1}{x} \cdot \left(-\frac{x^3}{2}\right) = -\frac{x^2}{2}$
Evaluate the Limit: $\lim_{x \to 0^+} \left(-\frac{x^2}{2}\right) = -\frac{0^2}{2} = 0$

Worked Example B: Indeterminate Difference via Common Denominator

Evaluate the limit:

\lim_{x \to 0} \left(\frac{1}{x} - \frac{1}{e^x - 1}\right)

Worked Solution:

Name the Governing Formula: Common Denominator Unification for Indeterminate Differences: $\frac{1}{f(x)} - \frac{1}{g(x)} = \frac{g(x) - f(x)}{f(x)g(x)}$ (This strategy is defined in Indeterminate Forms and L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- As $x \to 0$ , $\frac{1}{x} \to \pm\infty$ and $\frac{1}{e^0 - 1} = \frac{1}{0} \to \pm\infty$ .
- Indeterminate form: $\infty - \infty$ .
Combine Over a Common Denominator: $\frac{1}{x} - \frac{1}{e^x - 1} = \frac{(e^x - 1) - x}{x(e^x - 1)}$ Checking the form at $x = 0$ : Numerator is $e^0 - 1 - 0 = 0$ ; Denominator is $0(e^0 - 1) = 0$ . Form: $\frac{0}{0}$ . L'Hôpital's Rule applies.
First Application of L'Hôpital's Rule: Differentiate numerator and denominator (using Product Rule in denominator: $\frac{d}{dx}[x(e^x - 1)] = 1(e^x - 1) + x e^x$ ): $\lim_{x \to 0} \frac{e^x - 1}{e^x - 1 + xe^x}$ At $x = 0$ , this still yields $\frac{1 - 1}{1 - 1 + 0} = \frac{0}{0}$ . Apply L'Hôpital a second time.
Second Application of L'Hôpital's Rule: Differentiate numerator ( $e^x$ ) and denominator ( $e^x + (1\cdot e^x + xe^x) = 2e^x + xe^x$ ): $\lim_{x \to 0} \frac{e^x}{2e^x + xe^x} = \frac{e^0}{2e^0 + 0e^0} = \frac{1}{2(1) + 0} = \frac{1}{2}$

Practice Problem 5: Concept & Reading Check

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

(a) In the Conceptual Framework, explain why it is strategic to keep $\ln(x)$ in the numerator when rewriting an indeterminate product $0 \cdot \infty$ , rather than pushing $\ln(x)$ down to the denominator.
(b) Deconstruct Worked Example B: Why is it mathematically impossible to apply L'Hôpital's Rule directly to the two terms $\frac{1}{x} - \frac{1}{e^x - 1}$ separately? What role does the common denominator play?
(c) Review the linked Engineering Applications (Contaminant Dispersion Wellhead Flux (Environmental & Civil Engineering)). Why does the radial mass flux equation $J(r) = -D_0 r \ln(r / R_0)$ take an indeterminate form as $r \to 0^+$ , and how does resolving this limit prove the wellbore casing is safe from mineral clogging?

Solution Key:

(a) When converting an indeterminate product $f(x) \cdot g(x)$ where $f \to 0$ and $g = \ln(x) \to -\infty$ , inverting the algebraic polynomial factor $f(x) = x^n$ into the denominator as $x^{-n}$ produces a clean power whose derivative is simply $-n x^{-n-1}$ . When differentiated, the numerator becomes $\frac{1}{x}$ and the denominator is a power of $x$ , which multiply and cancel cleanly into an elementary monomial. In contrast, pushing $\ln(x)$ into the denominator creates $[\ln(x)]^{-1}$ . Differentiating this denominator yields $-[\ln(x)]^{-2} \cdot \frac{1}{x}$ , which introduces logarithmic powers into the denominator and creates a nested algebraic deadlock that prevents L'Hôpital from terminating.
(b) L'Hôpital's Rule is mathematically proven only for unified fractional limits of the form $\frac{f(x)}{g(x)}$ . It does not distribute across subtraction: $\lim (A - B) \neq \lim A' - \lim B'$ . The expression $\frac{1}{x} - \frac{1}{e^x - 1}$ represents two competing quantities that both approach infinity at comparable rates. Without combining them over a common denominator, there is no single numerator and denominator to differentiate. Unifying the fractions establishes a single composite fraction $\frac{e^x - 1 - x}{x(e^x - 1)}$ whose numerator and denominator both approach zero, creating the required $\frac{0}{0}$ structure.
(c) As groundwater converges toward the well centerline ( $r \to 0^+$ ), the radial setback distance vanishes ( $r \to 0$ ) while the logarithmic draw potential diverges ( $\ln(r / R_0) \to -\infty$ ). This creates a $0 \cdot (-\infty)$ indeterminate product. If the logarithmic potential dominated, the solute mass flux would explode to infinity at the borehole boundary, causing rapid chemical supersaturation, mineral scaling, and encrustation of the stainless-steel intake slots. By rewriting $J(r) = -D_0 \frac{\ln(r / R_0)}{r^{-1}}$ and applying L'Hôpital's Rule, environmental engineers prove that the linear distance convergence completely overpowers the logarithmic divergence, driving the flux smoothly to zero ( $\lim_{r \to 0^+} J(r) = 0$ ). This guarantees hydrodynamic stability and validates standard sand pack filtration sizing.

Show your steps and calculations here.

Practice Problem 6

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to 0^+} x \cot(\pi x)

Solution Key:

Name the Governing Formula: Trigonometric Reciprocal Inversion for Indeterminate Products: $x \cot(\pi x) = \frac{x}{\tan(\pi x)} = \frac{x \cos(\pi x)}{\sin(\pi x)}$ (This strategy is defined in Indeterminate Forms and L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- As $x \to 0^+$ , $x \to 0$ and $\cot(\pi x) \to \infty$ .
- Indeterminate form: $0 \cdot \infty$ .
Restructure as a Quotient: Using $\cot(\pi x) = \frac{1}{\tan(\pi x)}$ : $\lim_{x \to 0^+} x \cot(\pi x) = \lim_{x \to 0^+} \frac{x}{\tan(\pi x)}$ Checking form: at $x = 0$ , numerator is $0$ and $\tan(0) = 0$ . Form: $\frac{0}{0}$ . L'Hôpital's Rule applies.
Apply L'Hôpital's Rule: Differentiate numerator ( $\frac{d}{dx}[x] = 1$ ) and denominator ( $\frac{d}{dx}[\tan(\pi x)] = \pi\sec^2(\pi x)$ ): $\lim_{x \to 0^+} \frac{x}{\tan(\pi x)} \overset{\text{L'H}}{=} \lim_{x \to 0^+} \frac{1}{\pi\sec^2(\pi x)}$
Evaluate the Limit: Since $\sec(0) = \frac{1}{\cos(0)} = 1$ : $\frac{1}{\pi \sec^2(0)} = \frac{1}{\pi (1)^2} = \frac{1}{\pi}$

Show your steps and calculations here.

Practice Problem 7

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to \pi^-} \sin(x)\csc(3x)

Solution Key:

Name the Governing Formula: Reciprocal Trigonometric Quotient Inversion: $\sin(x)\csc(3x) = \frac{\sin(x)}{\sin(3x)}$ (This strategy is defined in Indeterminate Forms and L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- As $x \to \pi^-$ , $\sin(\pi) = 0$ and $\csc(3\pi) \to -\infty$ .
- Indeterminate form: $0 \cdot (-\infty)$ .
Form the Fractional Quotient: $\lim_{x \to \pi^-} \sin(x)\csc(3x) = \lim_{x \to \pi^-} \frac{\sin(x)}{\sin(3x)}$ Checking form: $\frac{\sin(\pi)}{\sin(3\pi)} = \frac{0}{0}$ . L'Hôpital's Rule applies.
Apply L'Hôpital's Rule: Differentiate numerator ( $\cos x$ ) and denominator via Chain Rule ( $3\cos(3x)$ ): $\lim_{x \to \pi^-} \frac{\sin(x)}{\sin(3x)} \overset{\text{L'H}}{=} \lim_{x \to \pi^-} \frac{\cos(x)}{3\cos(3x)}$
Evaluate the Determinate Limit: Substitute $x = \pi$ : $\frac{\cos(\pi)}{3\cos(3\pi)} = \frac{-1}{3(-1)} = \frac{-1}{-3} = \frac{1}{3}$

Show your steps and calculations here.

Practice Problem 8

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to 1^+} \left(\frac{5}{\ln x} - \frac{5}{x - 1}\right)

Solution Key:

Name the Governing Formula: Common Denominator Unification for Differences: $\frac{5}{\ln x} - \frac{5}{x - 1} = 5\left[\frac{(x - 1) - \ln x}{(x - 1)\ln x}\right]$ (This strategy is defined in Indeterminate Forms and L'Hôpital's Rule.)
Execute the STOP & CHECK Protocol:
- As $x \to 1^+$ , $\ln(x) \to 0^+ \implies \frac{5}{\ln x} \to +\infty$ .
- As $x \to 1^+$ , $x - 1 \to 0^+ \implies \frac{5}{x-1} \to +\infty$ .
- Indeterminate form: $\infty - \infty$ .
Combine Over a Common Denominator: $5 \lim_{x \to 1^+} \frac{x - 1 - \ln x}{(x - 1)\ln x}$ Checking form at $x = 1$ : Numerator is $1 - 1 - \ln(1) = 0$ ; Denominator is $(0)\ln(1) = 0$ . Form: $\frac{0}{0}$ .
First Application of L'Hôpital's Rule:
- Differentiate numerator: $\frac{d}{dx}[x - 1 - \ln x] = 1 - \frac{1}{x} = \frac{x - 1}{x}$ .
- Differentiate denominator via Product Rule: $\frac{d}{dx}[(x - 1)\ln x] = (1)\ln x + (x - 1)\frac{1}{x} = \ln x + 1 - \frac{1}{x} = \frac{x\ln x + x - 1}{x}$
- Dividing the two fractions cancels the common factor of $\frac{1}{x}$ : $5 \lim_{x \to 1^+} \frac{x - 1}{x\ln x + x - 1}$
Re-check Form:
- At $x = 1$ : Numerator is $1 - 1 = 0$ ; Denominator is $1(0) + 1 - 1 = 0$ . Form: still $\frac{0}{0}$ .
Second Application of L'Hôpital's Rule:
- Differentiate numerator: $\frac{d}{dx}[x - 1] = 1$ .
- Differentiate denominator via Product Rule on $x\ln x$ : $\frac{d}{dx}[x\ln x + x - 1] = \left(1\cdot\ln x + x\cdot\frac{1}{x}\right) + 1 - 0 = \ln x + 1 + 1 = \ln x + 2$
- Evaluate the limit at $x = 1$ : $5 \lim_{x \to 1^+} \frac{1}{\ln x + 2} = 5 \cdot \frac{1}{\ln(1) + 2} = 5 \cdot \frac{1}{0 + 2} = \frac{5}{2}$

Show your steps and calculations here.

Practice Problem 9

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to \infty} \left(\sqrt{x^2 + 6x} - x\right)

Solution Key:

Name the Governing Formula: Algebraic Conjugate Rationalization for Radical Differences: $\sqrt{A} - B = \frac{(\sqrt{A} - B)(\sqrt{A} + B)}{\sqrt{A} + B} = \frac{A - B^2}{\sqrt{A} + B}$ (This strategy is defined in Indeterminate Forms.)
Execute the STOP & CHECK Protocol:
- As $x \to \infty$ , $\sqrt{x^2 + 6x} \to \infty$ and $x \to \infty$ .
- Indeterminate form: $\infty - \infty$ .
Method 1: Conjugate Rationalization: Multiply and divide by the conjugate $\sqrt{x^2 + 6x} + x$ : $\lim_{x \to \infty} \frac{\left(\sqrt{x^2 + 6x} - x\right)\left(\sqrt{x^2 + 6x} + x\right)}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{6x}{\sqrt{x^2 + 6x} + x}$ Divide numerator and denominator by $x = \sqrt{x^2}$ for $x > 0$ : $\lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1} = \frac{6}{\sqrt{1 + 0} + 1} = \frac{6}{1 + 1} = \frac{6}{2} = 3$
Method 2: L'Hôpital's Rule Verification via Factorization: Factor out $x$ : $\lim_{x \to \infty} x\left(\sqrt{1 + \frac{6}{x}} - 1\right) = \lim_{x \to \infty} \frac{\left(1 + 6x^{-1}\right)^{1/2} - 1}{x^{-1}}$ Form is $\frac{0}{0}$ . Differentiating numerator and denominator: $\lim_{x \to \infty} \frac{\frac{1}{2}\left(1 + 6x^{-1}\right)^{-1/2} \cdot \left(-6x^{-2}\right)}{-x^{-2}} = \lim_{x \to \infty} \frac{3}{\sqrt{1 + 6x^{-1}}} = \frac{3}{\sqrt{1 + 0}} = 3$

Show your steps and calculations here.

Regroup 2

The Reciprocal Choice Strategy: When dealing with $0 \cdot \infty$ , ask: which term becomes simpler when inverted? Polynomial terms $x^n$ invert cleanly to $x^{-n}$ , whose derivative remains a power of $x$ . Inverting logarithms creates transcendental fractions that spiral out of control.
Algebraic Hygiene in Common Denominators: Combining fractions often produces terms with $\frac{1}{x}$ . Always multiply the numerator and denominator by $x$ to clear nested fractions before computing the next derivative.
Conjugate vs. L'Hôpital: In radical differences ( $\sqrt{x^2+ax}-x$ ), conjugate multiplication is often faster than L'Hôpital, but both provide mutual mathematical verification.

Skill Block 3

Worked Example A: Indeterminate Power of Type $1^\infty$

Evaluate the limit:

\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^{3x}

Worked Solution:

Name the Governing Formula: The 4-Step Logarithmic Transformation Algorithm: $\lim_{x \to a} y = e^L \quad \text{where } L = \lim_{x \to a} \ln y$ (This algorithm is defined in Indeterminate Powers and Logarithmic Transformation.)
Execute the STOP & CHECK Protocol:
- As $x \to \infty$ , $1 + \frac{2}{x} \to 1$ and $3x \to \infty$ .
- Indeterminate form: $1^\infty$ .
Step 1 & 2: Assign $y$ and Take the Natural Logarithm: Let $y = \left(1 + \frac{2}{x}\right)^{3x}$ . $\ln y = \ln\left[\left(1 + \frac{2}{x}\right)^{3x}\right] = 3x \ln\left(1 + \frac{2}{x}\right)$ As $x \to \infty$ , this product has form $\infty \cdot \ln(1) = \infty \cdot 0$ .
Step 3: Convert to a Quotient and Apply L'Hôpital's Rule: Push $x$ into the denominator as $x^{-1}$ , keeping constant $3$ outside: $L = \lim_{x \to \infty} \ln y = 3 \lim_{x \to \infty} \frac{\ln\left(1 + 2x^{-1}\right)}{x^{-1}}$ Checking form: $\frac{\ln(1)}{0} = \frac{0}{0}$ . L'Hôpital's Rule applies. Differentiating numerator via Chain Rule and denominator via Power Rule: $L = 3 \lim_{x \to \infty} \frac{\frac{1}{1 + 2x^{-1}} \cdot \left(-2x^{-2}\right)}{-x^{-2}}$ Canceling the common factor of $-x^{-2}$ : $L = 3 \lim_{x \to \infty} \frac{-2}{-1} \cdot \frac{1}{1 + 2/x} = 3 \cdot 2 \cdot \frac{1}{1 + 0} = 6$
Step 4: Exponentiate to Avoid the Halfway Trap: The limit of $\ln y$ is $L = 6$ . The limit of the original function $y$ is: $\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^{3x} = e^L = e^6$

Practice Problem 10: Concept & Reading Check

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

(a) In the Conceptual Framework, explain the mathematical tension (tug-of-war) present in the three indeterminate power forms ( $1^\infty, 0^0, \infty^0$ ).
(b) Deconstruct Worked Example A: Identify the "Halfway Trap" in Step 4. What would happen if a student reported $6$ as their final answer instead of $e^6$ ?
(c) Review the linked Engineering Applications (Continuous Intravenous Drug Infusion (Biomedical Engineering)). How does the discrete drug dosing model $C_n(t) = C_0 (1 + kt/n)^{-n}$ produce the indeterminate form $1^\infty$ as pulse frequency $n \to \infty$ , and what continuous function does it converge to?

Solution Key:

(a) Each indeterminate power represents a conflict between base and exponent properties:
- For $1^\infty$ : Any finite number strictly greater than $1$ raised to $\infty$ explodes to $\infty$ , while any number less than $1$ collapses to $0$ . The base approaches $1$ (which stays $1$ under any finite power), while the exponent approaches $\infty$ . The speed at which the base approaches $1$ relative to how fast the exponent diverges determines whether the expression converges to a finite positive constant, collapses to $0$ , or blows up to $\infty$ .
- For $0^0$ : A base approaching $0$ attempts to pull the expression to $0$ ( $0^k = 0$ ), whereas an exponent approaching $0$ attempts to pull any positive quantity to $1$ ( $c^0 = 1$ ).
- For $\infty^0$ : A base diverging to $\infty$ pulls toward $\infty$ , while a vanishing exponent pulls toward $1$ .
(b) The Halfway Trap occurs when a student successfully computes the limit of the natural logarithm $L = \lim \ln y = 6$ , and mistakenly writes $6$ on their exam sheet. $6$ is the asymptotic exponent, not the function value. Because the natural logarithm was introduced in Step 2 to pull down the exponent, the original function is related by $y = e^{\ln y}$ . The final answer must be exponentiated: $\lim y = e^6 \approx 403.43$ . Leaving the answer as $6$ creates a massive mathematical and physical error ( $403.4$ vs $6$ ).
(c) When automated infusion pumps deliver medication in $n$ discrete micro-boluses per hour, the remaining drug concentration is $C_n(t) = C_0 \left(1 + \frac{kt}{n}\right)^{-n}$ . In the limit of continuous steady infusion, pulse frequency $n \to \infty$ . The term $\frac{kt}{n} \to 0$ , making the base approach $1$ , while the exponent $-n \to -\infty$ , producing the indeterminate power form $1^{-\infty} = 1^\infty$ . Applying the logarithmic transformation algorithm and L'Hôpital's Rule on $\ln y = \frac{-\ln(1 + kt n^{-1})}{n^{-1}}$ proves that $L = -kt$ , so the limit exponentiates to $e^{-kt}$ . Multiplying by $C_0$ yields the continuous exponential decay model $C(t) = C_0 e^{-kt}$ , guaranteeing that high-frequency micro-pulsing smoothly mimics continuous biological drug clearance without concentration discontinuities.

Show your steps and calculations here.

Practice Problem 11

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to \infty} \left(1 - \frac{8}{x}\right)^{5x}

Solution Key:

Name the Governing Formula: The 4-Step Logarithmic Transformation Algorithm: $\lim_{x \to a} y = e^L \quad \text{where } L = \lim_{x \to a} \ln y$ (This algorithm is defined in Indeterminate Powers and Logarithmic Transformation.)
Execute the STOP & CHECK Protocol:
- As $x \to \infty$ , $1 - \frac{8}{x} \to 1$ and $5x \to \infty$ .
- Indeterminate form: $1^\infty$ .
Step 1 & 2: Assign $y$ and Take the Natural Logarithm: Let $y = \left(1 - \frac{8}{x}\right)^{5x}$ . $\ln y = 5x \ln\left(1 - \frac{8}{x}\right)$ Form: $\infty \cdot \ln(1) = \infty \cdot 0$ .
Step 3: Convert to a Quotient and Apply L'Hôpital's Rule: Push $x$ into the denominator as $x^{-1}$ , keeping $5$ in front: $L = \lim_{x \to \infty} \ln y = 5 \lim_{x \to \infty} \frac{\ln\left(1 - 8x^{-1}\right)}{x^{-1}}$ Checking form: $\frac{\ln(1)}{0} = \frac{0}{0}$ . L'Hôpital's Rule applies. Differentiate numerator and denominator with respect to $x$ : $L = 5 \lim_{x \to \infty} \frac{\frac{1}{1 - 8x^{-1}} \cdot \left(8x^{-2}\right)}{-x^{-2}}$ Cancel common $x^{-2}$ terms: $L = 5 \lim_{x \to \infty} \frac{8}{-1} \cdot \frac{1}{1 - 8/x} = 5 \cdot (-8) \cdot \frac{1}{1 - 0} = -40$
Step 4: Exponentiate to Recover the Original Limit: $\lim_{x \to \infty} \left(1 - \frac{8}{x}\right)^{5x} = e^L = e^{-40} = \frac{1}{e^{40}}$

Show your steps and calculations here.

Practice Problem 12

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to \infty} \left(x^2 - 1\right)^{2/x}

Solution Key:

Name the Governing Formula: The 4-Step Logarithmic Transformation Algorithm: $\lim_{x \to a} y = e^L \quad \text{where } L = \lim_{x \to a} \ln y$ (This algorithm is defined in Indeterminate Powers and Logarithmic Transformation.)
Execute the STOP & CHECK Protocol:
- As $x \to \infty$ , $x^2 - 1 \to \infty$ and $\frac{2}{x} \to 0$ .
- Indeterminate form: $\infty^0$ .
Step 1 & 2: Assign $y$ and Take the Natural Logarithm: Let $y = \left(x^2 - 1\right)^{2/x}$ . $\ln y = \frac{2}{x} \ln\left(x^2 - 1\right) = \frac{2\ln\left(x^2 - 1\right)}{x}$
Step 3: Evaluate the Limit of $\ln y$ : As $x \to \infty$ , numerator $2\ln(\infty) \to \infty$ and denominator $x \to \infty$ . Form: $\frac{\infty}{\infty}$ . L'Hôpital's Rule applies. Differentiate numerator via Chain Rule and denominator via Power Rule: $L = \lim_{x \to \infty} \frac{\frac{d}{dx}[2\ln(x^2 - 1)]}{\frac{d}{dx}[x]} = \lim_{x \to \infty} \frac{2 \cdot \frac{2x}{x^2 - 1}}{1} = \lim_{x \to \infty} \frac{4x}{x^2 - 1}$ Form is still $\frac{\infty}{\infty}$ . Divide numerator and denominator by $x^2$ (or apply L'Hôpital a second time): $L = \lim_{x \to \infty} \frac{\frac{4}{x}}{1 - \frac{1}{x^2}} = \frac{0}{1 - 0} = 0$
Step 4: Exponentiate to Recover the Original Limit: Since $L = 0$ : $\lim_{x \to \infty} \left(x^2 - 1\right)^{2/x} = e^L = e^0 = 1$

Show your steps and calculations here.

Practice Problem 13

Evaluate the following limit using appropriate methods. Be sure to indicate the indeterminate form and each time L'Hôpital's Rule is used:

\lim_{x \to 0^+} x^{\sin x}

Solution Key:

Name the Governing Formula: Logarithmic Transformation for Indeterminate Powers of Type $0^0$ : $\lim_{x \to a} y = e^L \quad \text{where } L = \lim_{x \to a} \ln y$ (This algorithm is defined in Indeterminate Powers and Logarithmic Transformation.)
Execute the STOP & CHECK Protocol:
- As $x \to 0^+$ , base $x \to 0^+$ and exponent $\sin(x) \to 0$ .
- Indeterminate form: $0^0$ .
Step 1 & 2: Assign $y$ and Take the Natural Logarithm: Let $y = x^{\sin x}$ . $\ln y = \sin(x) \ln(x)$ Form: $0 \cdot (-\infty)$ .
Step 3: Convert to a Quotient and Apply L'Hôpital's Rule: Rewrite $\sin(x)$ as $\frac{1}{\csc(x)}$ in the denominator: $L = \lim_{x \to 0^+} \frac{\ln(x)}{\csc(x)}$ Checking form: as $x \to 0^+$ , $\ln(x) \to -\infty$ and $\csc(x) \to +\infty$ . Form: $\frac{-\infty}{\infty}$ . Differentiate numerator ( $\frac{1}{x}$ ) and denominator ( $-\csc x \cot x$ ): $L = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\csc(x)\cot(x)} = \lim_{x \to 0^+} \left[-\frac{\sin(x)\tan(x)}{x}\right]$ Split into known fundamental limits: $L = -\lim_{x \to 0^+} \left[\frac{\sin(x)}{x}\right] \cdot \lim_{x \to 0^+} [\tan(x)] = -(1) \cdot (0) = 0$
Step 4: Exponentiate to Recover the Original Limit: Since $L = 0$ : $\lim_{x \to 0^+} x^{\sin x} = e^L = e^0 = 1$

Show your steps and calculations here.

Regroup 3

Avoiding the Halfway Trap: Re-drill with students that whenever they see an exponent in the original limit ( $f(x)^{g(x)}$ ), the very last line of their worksheet solution must contain an $e^{(\cdot)}$ !
Asymptotic Power Comparisons: Notice how in Practice Problems 12 and 13, despite different bases ( $\infty$ vs $0$ ), the vanishing exponent pulls both limits to $e^0 = 1$ .
Synthesizing the Entire Indeterminate Taxonomy: Review how products, differences, and powers all funnel into the central engine of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before L'Hôpital can be engaged.

Synthesis Wrap-up

The Universal L'Hôpital Decision Protocol:
1. Direct Substitution: Always test $x = a$ first. If the form is determinate, evaluate immediately. Never differentiate a determinate limit!
2. Direct Quotients ( $\frac{0}{0}, \frac{\pm\infty}{\pm\infty}$ ): Apply L'Hôpital's Rule immediately by differentiating numerator and denominator separately.
3. Products ( $0 \cdot \pm\infty$ ): Keep logarithms in the numerator; push polynomial/algebraic terms to the denominator via reciprocal inversion ( $x \to x^{-1}$ ).
4. Differences ( $\infty - \infty$ ): Combine fractions over a common denominator or multiply by algebraic conjugates.
5. Powers ( $1^\infty, 0^0, \infty^0$ ): Set $y = [f(x)]^{g(x)}$ , evaluate $L = \lim \ln y$ , and compute the final answer as $e^L$ .
Homework Assignment: Complete the Knewton Alta assignment: L'Hospital's Rule. Practice writing down the formal indeterminate form above the equals sign ( $\overset{\text{L'H}}{=}$ ) every time you apply the rule on your homework scratch sheets.

MATH-181 (Calculus I) Curriculum

Session 12 - L’Hopital’s Rule and Indeterminate Forms

Micro-Lecture

Skill Block 1

Worked Example A: Indeterminate Quotient with Chain Rule

Worked Example B: Repeated L'Hôpital on an Exponential Quotient

Practice Problem 1: Concept & Reading Check

Practice Problem 2

Practice Problem 3

Practice Problem 4

Regroup 1

Skill Block 2

Worked Example A: Indeterminate Product Inversion

Worked Example B: Indeterminate Difference via Common Denominator

Practice Problem 5: Concept & Reading Check

Practice Problem 6

Practice Problem 7

Practice Problem 8

Practice Problem 9

Regroup 2

Skill Block 3

Worked Example A: Indeterminate Power of Type 1∞1^\infty1∞

Practice Problem 10: Concept & Reading Check

Practice Problem 11

Practice Problem 12

Practice Problem 13

Regroup 3

Synthesis Wrap-up

Worked Example A: Indeterminate Power of Type $1^\infty$