4.6 L'Hôpital's Rule and Indeterminate Forms

4.5 L'Hôpital's Rule and Indeterminate Forms

Quick Answer: L'Hôpital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ yields the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ provided the derivative limit exists or is $\pm\infty$. Never use the Quotient Rule for L'Hôpital's Rule—differentiate the top and bottom separately. Other indeterminate forms ($0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, $\infty^0$) must first be algebraically transformed into $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

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1. Indeterminate Quotients and Theorem Statement

When evaluating limits by direct substitution, an expression like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ represents an indeterminate form—the limit cannot be determined without deeper mathematical analysis because competing forces are at play.

Formal Theorem Statement

Let $f$ and $g$ be differentiable functions on an open interval containing $c$ (except possibly at $c$ itself), where $g'(x) \neq 0$ near $c$. If: $\lim_{x \to c} f(x) = 0 \quad \text{and} \quad \lim_{x \to c} g(x) = 0$ $\text{OR} \quad \lim_{x \to c} f(x) = \pm\infty \quad \text{and} \quad \lim_{x \to c} g(x) = \pm\infty$ then: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$ provided this limit exists as a finite real number or is $\pm\infty$. The theorem holds equally for one-sided limits ($x \to c^+$, $x \to c^-$) and limits at infinity ($x \to \infty$, $x \to -\infty$).

                          L'Hôpital's Rule Decision Tree

                              Evaluate: lim f(x)/g(x)
                                         │
                     ┌───────────────────┴───────────────────┐
                     ▼                                       ▼
          Direct Substitution yields:             Direct Substitution yields:
             0/0  or  ±∞/±∞                          Determinate Form (e.g., 5/2, 0/5, 1/0)
                     │                                       │
                     ▼                                       ▼
          APPLY L'HÔPITAL'S RULE:                 DO NOT APPLY L'HÔPITAL!
             Evaluate: lim f'(x)/g'(x)               State answer directly or evaluate asymptote.

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2. The Cardinal Rule: Separate Differentiation

CRITICAL WARNING: Applying L'Hôpital's Rule means taking the independent derivative of the numerator $f'(x)$ and dividing it by the independent derivative of the denominator $g'(x)$: $\text{Correct:} \quad \lim \frac{f'(x)}{g'(x)}$ $\text{FATAL ERROR:} \quad \lim \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \quad (\text{Quotient Rule is NOT used here!})$

Basic Example: Direct Application

Evaluate $\lim_{x \to 0} \frac{\sin(3x)}{x}$.

1. Check form: Direct substitution gives $\frac{\sin 0}{0} = \frac{0}{0}$ (Indeterminate quotient).
2. Apply L'Hôpital's Rule: $\lim_{x \to 0} \frac{\frac{d}{dx}[\sin(3x)]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{3\cos(3x)}{1} = \frac{3\cos(0)}{1} = \frac{3(1)}{1} = 3$

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3. Repeated Applications and Dominance Hierarchies

If the derivative ratio $\frac{f'(x)}{g'(x)}$ still yields $\frac{0}{0}$ or $\frac{\infty}{\infty}$, verify that the conditions hold and apply L'Hôpital's Rule a second (or third) time.

Example: Two-Stage L'Hôpital Evaluation

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

1. First test: $\frac{e^0 - 0 - 1}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}$. (Apply L'Hôpital #1): $\lim_{x \to 0} \frac{e^x - 1}{2x}$
2. Second test: $\frac{e^0 - 1}{2(0)} = \frac{0}{0}$. (Apply L'Hôpital #2): $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$

Growth Rate Dominance Hierarchy at Infinity

Repeated application of L'Hôpital's Rule establishes the fundamental asymptotic growth rate hierarchy as $x \to \infty$:

$\ln x \ll x^p \ll a^x \ll x! \ll x^x \quad (p > 0, a > 1)$

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4. Transforming Other Indeterminate Forms

L'Hôpital's Rule cannot be directly applied to products, differences, or powers. You must convert them to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first.

Type 1: Indeterminate Products ($0 \cdot \infty$)

Rewrite $f(x) \cdot g(x)$ as a fraction by inverting one factor: $f(x) \cdot g(x) = \frac{f(x)}{1/g(x)} \;\;\left(\frac{0}{0}\right) \qquad \text{or} \qquad f(x) \cdot g(x) = \frac{g(x)}{1/f(x)} \;\;\left(\frac{\infty}{\infty}\right)$

Worked Example: Evaluate $\lim_{x \to 0^+} x \ln x$.

1. Direct substitution gives $0 \cdot (-\infty)$ (Indeterminate product).
2. Rewrite with $x$ moved to the denominator as $x^{-1}$: $\lim_{x \to 0^+} \frac{\ln x}{1/x} \quad \left(\frac{-\infty}{\infty}\right)$
3. Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \left( \frac{1}{x} \cdot (-x^2) \right) = \lim_{x \to 0^+} (-x) = 0$

Type 2: Indeterminate Differences ($\infty - \infty$)

Combine rational expressions over a common denominator or multiply by algebraic conjugates.

Worked Example: Evaluate $\lim_{x \to 1} \left( \frac{1}{\ln x} - \frac{1}{x - 1} \right)$.

1. Form is $\infty - \infty$.
2. Combine into a single rational fraction: $\lim_{x \to 1} \frac{(x - 1) - \ln x}{(x - 1)\ln x} \quad \left(\frac{0}{0}\right)$
3. Apply L'Hôpital's Rule #1: $\lim_{x \to 1} \frac{1 - 1/x}{(1)\ln x + (x - 1)(1/x)} = \lim_{x \to 1} \frac{\frac{x - 1}{x}}{\ln x + 1 - 1/x} = \lim_{x \to 1} \frac{x - 1}{x \ln x + x - 1} \quad \left(\frac{0}{0}\right)$
4. Apply L'Hôpital's Rule #2: $\lim_{x \to 1} \frac{1}{\ln x + x(1/x) + 1} = \lim_{x \to 1} \frac{1}{\ln x + 2} = \frac{1}{0 + 2} = \frac{1}{2}$

Type 3: Indeterminate Powers ($1^\infty$, $0^0$, $\infty^0$)

Use the 3-Step Logarithmic Limit Method:

1. Let $y = [f(x)]^{g(x)}$ and take the natural log: $\ln y = g(x) \ln(f(x))$.
2. Compute the limit of the logarithm: $L = \lim_{x \to c} [g(x) \ln(f(x))]$ (using product transformation and L'Hôpital).
3. Exponentiate to recover the original limit: $\lim_{x \to c} y = e^L$.

Worked Example: Evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x}$.

1. Form is $1^\infty$.
2. Let $y = \left(1 + \frac{3}{x}\right)^{2x} \implies \ln y = 2x \ln\left(1 + \frac{3}{x}\right) = \frac{2\ln(1 + 3/x)}{1/x}$.
3. Evaluate $L = \lim_{x \to \infty} \frac{2\ln(1 + 3/x)}{1/x}$ (form $\frac{0}{0}$): $L = \lim_{x \to \infty} \frac{2 \cdot \frac{-3/x^2}{1 + 3/x}}{-1/x^2} = \lim_{x \to \infty} \frac{6}{1 + 3/x} = \frac{6}{1 + 0} = 6$
4. Exponentiate: $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x} = e^L = e^6$.

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5. Non-Indeterminate (Determinate) Forms

Applying L'Hôpital's Rule to determinate forms produces completely false results. Always test direct substitution before differentiating!

Caution Example

Evaluate $\lim_{x \to 0} \frac{x + 2}{x + 3}$.

• Direct substitution gives $\frac{0 + 2}{0 + 3} = \frac{2}{3}$ (Correct answer!).
• If you mistakenly applied L'Hôpital's Rule: $\lim_{x \to 0} \frac{1}{1} = 1$ (Completely WRONG!).

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6. CLEP Traps and Exam Strategies

• Trap 1: Applying L'Hôpital's Rule blindly without checking $0/0$ or $\infty/\infty$. This is the number one trap engineered by exam writers.
• Trap 2: Forgetting to exponentiate ($e^L$) at the end of indeterminate power problems. Remember $L$ is only the limit of $\ln y$. Your final answer must be $e^L$.
• Trap 3: Using the Quotient Rule. Never write $\frac{f'g - fg'}{g^2}$ during L'Hôpital's Rule. Take derivative of top, derivative of bottom.