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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.
| Property | Value |
|---|---|
| Concept | Indeterminate Forms & L'Hospital's Rule |
| Chapter | Chapter 6, Section 8 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
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.
If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, try:
$$\boxed{f - g = \frac{\frac{1}{g} - \frac{1}{f}}{\frac{1}{fg}} \quad \text{(common denominator approach)}}$$
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
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: $\lim_{x \to 0^+} x \ln x = 0$
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.
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: $\lim_{x \to (\pi/2)^-} (\sec x - \tan x) = 0$
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: $\lim_{x \to \infty} (e^x - x) = \infty$
Evaluate $\lim_{x \to \infty} x \cdot e^{-x}$
Evaluate $\lim_{x \to 0^+} \sqrt{x} \cdot \cot x$
Evaluate $\lim_{x \to 1} \left(\frac{1}{\ln x} - \frac{1}{x - 1}\right)$
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
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 $\lim_{x \to \infty} x^{-p} \ln x$.
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:
The Goal: Transform into $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then let L'Hospital do the work.
Looking back:
Looking ahead:
Real-world connections:
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|---|---|---|
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Last updated: 2026-01-22