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L'Hospital's Rule for Products and Differences

MATH161
Reference: Stewart 6.8  •  Chapter: 6  •  Section: 8

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L'Hospital's Rule for Products and Differences

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

This skill
L'Hospital's Rule (Products & Differences)

Quick Reference

Property Value
Concept Indeterminate Forms & L'Hospital's Rule
Chapter Chapter 6, Section 8
Difficulty Intermediate
Time ~20 minutes

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:

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

Rewrite Resulting Form When to Use
$\frac{f}{1/g}$ $\frac{0}{0}$ When differentiating $1/g$ is simpler
$\frac{g}{1/f}$ $\frac{\infty}{\infty}$ When differentiating $1/f$ is simpler

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

Step 2: Convert to quotient

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

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

Let's 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

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

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

Level 1 Basic Product Conversion

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

Thought Process
  1. Identify form: As $x \to \infty$: $x \to \infty$ and $e^{-x} \to 0$. Form: $\infty \cdot 0$ ✓
  1. Convert: Write as $\frac{x}{e^x}$ (form $\frac{\infty}{\infty}$)
  1. Apply L'Hospital: $\frac{d}{dx}[x] = 1$, $\frac{d}{dx}[e^x] = e^x$
Show Answer

Identify form: $\infty \cdot 0$ ✓

Convert: $x \cdot e^{-x} = \frac{x}{e^x}$

Apply L'Hospital: $$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = \boxed{0}$$

Level 2 Product with Trig

Evaluate $\lim_{x \to 0^+} \sqrt{x} \cdot \cot x$

Thought Process
  1. Identify form: $\sqrt{x} \to 0^+$ and $\cot x = \frac{\cos x}{\sin x} \to \frac{1}{0^+} = +\infty$. Form: $0 \cdot \infty$ ✓
  1. Convert: $\sqrt{x} \cdot \cot x = \sqrt{x} \cdot \frac{\cos x}{\sin x} = \frac{\sqrt{x} \cos x}{\sin x}$
  1. Check form: As $x \to 0^+$: numerator $\to 0$, denominator $\to 0$. Form: $\frac{0}{0}$ ✓
  1. Apply L'Hospital (using product rule on numerator)
Show Answer

Identify form: $0 \cdot \infty$ ✓

Rewrite strategically: $$\sqrt{x} \cot x = \frac{\sqrt{x} \cos x}{\sin x} = \frac{\sqrt{x}}{\sin x} \cdot \cos x$$

Analyze each factor separately:

Since $\lim_{x \to 0^+} \cos x = 1$, we focus on $\frac{\sqrt{x}}{\sin x}$.

Rewrite using a known limit: $$\frac{\sqrt{x}}{\sin x} = \frac{\sqrt{x}}{x} \cdot \frac{x}{\sin x} = \frac{1}{\sqrt{x}} \cdot \frac{x}{\sin x}$$

As $x \to 0^+$:

  • $\frac{x}{\sin x} \to 1$ (standard limit)
  • $\frac{1}{\sqrt{x}} \to +\infty$

Combine: $$\lim_{x \to 0^+} \sqrt{x} \cot x = \lim_{x \to 0^+} \left(\frac{1}{\sqrt{x}} \cdot \frac{x}{\sin x} \cdot \cos x\right) = (+\infty)(1)(1) = +\infty$$

Therefore: $\boxed{\lim_{x \to 0^+} \sqrt{x} \cot x = +\infty}$

Key insight: Sometimes rewriting and using known limits is cleaner than forcing L'Hospital's Rule.

Level 3 Difference Using Common Denominator

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

Thought Process
  1. Identify form: As $x \to 1$: $\ln x \to 0$ so $\frac{1}{\ln x} \to \pm\infty$; $x - 1 \to 0$ so $\frac{1}{x-1} \to \pm\infty$. Form: $\infty - \infty$ ✓
  1. Common denominator:
  1. Check form: Numerator and denominator both $\to 0$. Form: $\frac{0}{0}$ ✓
  1. Apply L'Hospital
Show Answer

Identify form: $\infty - \infty$ ✓

Common denominator: $$\frac{1}{\ln x} - \frac{1}{x-1} = \frac{(x-1) - \ln x}{(x-1)\ln x}$$

Check form: At $x = 1$:

  • Numerator: $(1-1) - \ln 1 = 0 - 0 = 0$
  • Denominator: $(0)(\ln 1) = 0$
  • Form: $\frac{0}{0}$ ✓

Apply L'Hospital: $$\lim_{x \to 1} \frac{(x-1) - \ln x}{(x-1)\ln x} = \lim_{x \to 1} \frac{1 - 1/x}{\ln x + (x-1)/x}$$

Simplify: $\frac{1 - 1/x}{\ln x + (x-1)/x} = \frac{(x-1)/x}{\ln x + (x-1)/x}$

At $x = 1$: Still $\frac{0}{0}$. Apply L'Hospital again:

Numerator derivative: $\frac{d}{dx}\left[\frac{x-1}{x}\right] = \frac{d}{dx}\left[1 - \frac{1}{x}\right] = \frac{1}{x^2}$

Denominator derivative: $\frac{1}{x} + \frac{x - (x-1)}{x^2} = \frac{1}{x} + \frac{1}{x^2}$

$$= \lim_{x \to 1} \frac{1/x^2}{1/x + 1/x^2} = \frac{1}{1 + 1} = \boxed{\frac{1}{2}}$$

Level 4 Curve Sketching Application

For the function $f(x) = x e^{-x}$:

(a) Find $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$

(b) Use these limits to identify any horizontal asymptotes

(c) Find the critical point and determine if it's a max or min

Thought Process

(a) For $x \to \infty$: Form is $\infty \cdot 0$, convert to $\frac{x}{e^x}$ and use L'Hospital. For $x \to -\infty$: $e^{-x} \to e^{+\infty} = \infty$ and $x \to -\infty$, so $f(x) \to -\infty$.

(b) Horizontal asymptote is where $\lim$ is finite.

(c) $f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1-x)$. Set to 0: $x = 1$ (since $e^{-x} \neq 0$).

Show Answer

(a) Limits:

As $x \to \infty$: Form: $\infty \cdot 0$. Convert: $\frac{x}{e^x}$.

$$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0$$

As $x \to -\infty$: $$\lim_{x \to -\infty} x e^{-x} = \lim_{x \to -\infty} \frac{x}{e^x}$$

As $x \to -\infty$: numerator $\to -\infty$, denominator $e^x \to 0^+$.

$$= \frac{-\infty}{0^+} = -\infty$$

(b) Horizontal asymptotes:

$y = 0$ is a horizontal asymptote as $x \to \infty$.

No horizontal asymptote as $x \to -\infty$.

(c) Critical points:

$$f'(x) = e^{-x} - xe^{-x} = e^{-x}(1 - x)$$

$f'(x) = 0$ when $1 - x = 0$, i.e., $x = 1$.

  • For $x < 1$: $f'(x) > 0$ (increasing)
  • For $x > 1$: $f'(x) < 0$ (decreasing)

So $x = 1$ is a local maximum.

$f(1) = 1 \cdot e^{-1} = \boxed{\frac{1}{e} \approx 0.368}$

Level 5 Proving Logarithm Grows Slower Than Any Power

Prove that for any $p > 0$:

$$\lim_{x \to \infty} \frac{\ln x}{x^p} = 0$$

This shows that $\ln x$ grows more slowly than any positive power of $x$.

Hint: Use the product form $\lim_{x \to \infty} x^{-p} \ln x$.

Thought Process

Strategy:

The limit $\frac{\ln x}{x^p}$ is form $\frac{\infty}{\infty}$ for any $p > 0$.

Apply L'Hospital directly: $$\frac{d}{dx}[\ln x] = \frac{1}{x}, \quad \frac{d}{dx}[x^p] = px^{p-1}$$

$$\lim_{x \to \infty} \frac{1/x}{px^{p-1}} = \lim_{x \to \infty} \frac{1}{px^p} = 0$$

The key insight: one application suffices because $\frac{1/x}{px^{p-1}} = \frac{1}{px^p}$, which is a simple power that goes to 0.

Show Answer

Proof:

Let $p > 0$ be arbitrary. Consider $L = \lim_{x \to \infty} \frac{\ln x}{x^p}$.

Step 1: Verify form is $\frac{\infty}{\infty}$ ✓

Step 2: Apply L'Hospital's Rule:

$$L = \lim_{x \to \infty} \frac{\frac{d}{dx}[\ln x]}{\frac{d}{dx}[x^p]} = \lim_{x \to \infty} \frac{1/x}{px^{p-1}}$$

Step 3: Simplify:

$$L = \lim_{x \to \infty} \frac{1}{px \cdot x^{p-1}} = \lim_{x \to \infty} \frac{1}{px^p}$$

Step 4: Evaluate:

Since $p > 0$, we have $x^p \to \infty$ as $x \to \infty$.

Therefore: $$L = \frac{1}{\infty} = 0$$

Conclusion: $$\boxed{\lim_{x \to \infty} \frac{\ln x}{x^p} = 0 \text{ for all } p > 0}$$

Interpretation: No matter how small the power $p$ (e.g., $p = 0.001$), eventually $x^p$ outgrows $\ln x$. The logarithm is the "slowest" growing unbounded elementary function. $\square$

Mastery Checklist

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?"
  1. Differences ($\infty - \infty$): Think "How can I combine these?"

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


Connections

Looking back:

Looking ahead:

Real-world connections:


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Last updated: 2026-01-22