L'Hospital's Rule for Products and Differences

5 min read

Converting the Unconvertible

L’Hospital’s Rule directly handles $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms. But what about $0 \cdot \infty$ or $\infty - \infty$?

The key insight: these forms can be rewritten as quotients, bringing them into L’Hospital territory.

Consider $\lim_{x \to 0^+} x \ln x$. As $x \to 0^+$, we have $x \to 0$ and $\ln x \to -\infty$. That’s form $0 \cdot (-\infty)$, a tug-of-war between something shrinking and something exploding. Who wins?

By rewriting cleverly, we’ll find out.

Prerequisite Map

Quick Reference

Key Concepts

Indeterminate Products: $0 \cdot \infty$

If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = \pm\infty$, rewrite:

$$\boxed{f \cdot g = \frac{f}{1/g} \quad \text{or} \quad f \cdot g = \frac{g}{1/f}}$$

Strategy: Choose the form that leads to simpler derivatives.

Indeterminate Differences: $\infty - \infty$

If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, try:

1. Common denominator: Combine fractions
2. Factor out: Extract dominant term
3. Rationalize: Multiply by conjugate

$$\boxed{f - g = \frac{\frac{1}{g} - \frac{1}{f}}{\frac{1}{fg}} \quad \text{(common denominator approach)}}$$

Decision Guide

INDETERMINATE PRODUCT (0 · ∞)
─────────────────────────────────
            ↓
    f → 0 and g → ∞
            ↓
    ┌───────────────────┐
    │ Which is easier?  │
    └───────────────────┘
           /    \
          /      \
    1/g simpler   1/f simpler
         ↓              ↓
    Write f/(1/g)   Write g/(1/f)
         ↓              ↓
    Form: 0/0      Form: ∞/∞
         ↓              ↓
    Apply L'Hospital's Rule


INDETERMINATE DIFFERENCE (∞ - ∞)
─────────────────────────────────
            ↓
    f → ∞ and g → ∞
            ↓
    ┌─────────────────────────┐
    │ What form are f and g?  │
    └─────────────────────────┘
       /         |          \
   Fractions  Radicals   Other
      ↓          ↓          ↓
   Common    Rationalize  Factor
   denom.    (conjugate)   out
      ↓          ↓          ↓
   New quotient form → L'Hospital

Worked Examples

Example 1: Product Form $0 \cdot \infty$

Evaluate $\lim_{x \to 0^+} x \ln x$

Step 1: Identify form

• $x \to 0^+$
• $\ln x \to -\infty$
• Form: $0 \cdot (-\infty)$ ✓

Step 2: Convert to quotient

Option A: $\frac{\ln x}{1/x}$ gives $\frac{-\infty}{\infty}$ ✓

Option B: $\frac{x}{1/\ln x}$ gives $\frac{0}{0}$ ✓

Try Option A (usually cleaner for logarithms):

Step 3: Apply L’Hospital’s Rule $$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2}$$

Step 4: Simplify $$= \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0$$

Answer: $\lim_{x \to 0^+} x \ln x = 0$

Example 2: Why Choice Matters

Evaluate $\lim_{x \to 0^+} x \ln x$ using the OTHER conversion.

Using $\frac{x}{1/\ln x}$:

$$\lim_{x \to 0^+} \frac{x}{1/\ln x} = \lim_{x \to 0^+} \frac{1}{-1/(\ln x)^2 \cdot (1/x)}$$

This becomes: $$= \lim_{x \to 0^+} \frac{1}{-1/(x(\ln x)^2)} = \lim_{x \to 0^+} \left(-x(\ln x)^2\right)$$

Worse than where we started! This requires more work.

Lesson: Choose your conversion wisely.

Example 3: Difference Form $\infty - \infty$

Evaluate $\lim_{x \to (\pi/2)^-} (\sec x - \tan x)$

Step 1: Identify form

• $\sec x \to +\infty$ as $x \to (\pi/2)^-$
• $\tan x \to +\infty$ as $x \to (\pi/2)^-$
• Form: $\infty - \infty$ ✓

Step 2: Convert using common denominator $$\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$$

Step 3: Check new form

• Numerator: $1 - \sin(\pi/2) = 1 - 1 = 0$
• Denominator: $\cos(\pi/2) = 0$
• Form: $\frac{0}{0}$ ✓

Step 4: Apply L’Hospital’s Rule $$\lim_{x \to (\pi/2)^-} \frac{1 - \sin x}{\cos x} = \lim_{x \to (\pi/2)^-} \frac{-\cos x}{-\sin x} = \frac{0}{-1} = 0$$

Answer: $\lim_{x \to (\pi/2)^-} (\sec x - \tan x) = 0$

Example 4: Factoring Approach

Evaluate $\lim_{x \to \infty} (e^x - x)$

Step 1: Form is $\infty - \infty$

Step 2: Factor out the dominant term $$e^x - x = e^x\left(1 - \frac{x}{e^x}\right)$$

Step 3: Analyze $\lim_{x \to \infty} \frac{x}{e^x}$

Form: $\frac{\infty}{\infty}$. Apply L’Hospital: $$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0$$

Step 4: Combine $$\lim_{x \to \infty} e^x\left(1 - \frac{x}{e^x}\right) = \lim_{x \to \infty} e^x \cdot (1 - 0) = \infty$$

Answer: $\lim_{x \to \infty} (e^x - x) = \infty$

Practice Problems

Common Misconceptions

Mastery Checklist

• I can convert $0 \cdot \infty$ to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form
• I can choose the conversion that leads to simpler derivatives
• I can convert $\infty - \infty$ using common denominators
• I can factor out dominant terms when appropriate
• I recognize that $x e^{-x} \to 0$ as $x \to \infty$ (common pattern)
• I can apply these techniques to curve sketching problems

Mental Model

The “Rewrite and Conquer” Strategy:

L’Hospital’s Rule is a powerful weapon, but it only works on quotients. When you encounter products or differences:

1. Products ($0 \cdot \infty$): Think “How can I write this as a fraction?”

   • Move one factor to the denominator by taking its reciprocal
   • Choose the reciprocal that’s easier to differentiate

2. Differences ($\infty - \infty$): Think “How can I combine these?”

   • Common denominator if dealing with fractions
   • Factoring if one term dominates
   • Conjugate multiplication for roots

The Goal: Transform into $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then let L’Hospital do the work.

Connections

Looking back:

• Recognizing Indeterminate Forms: identify the form first
• L’Hospital’s Rule (Quotients): the core technique you’re converting TO

Looking ahead:

• L’Hospital’s Rule (Powers): similar conversion idea using logarithms
• Curve Sketching: often need limits at infinity

Real-world connections:

• In probability: $\lim_{x \to \infty} x e^{-x}$ appears in exponential distribution analysis
• In physics: Products of growth and decay terms model damped oscillations

Last updated: 2026-01-22