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L'Hospital's Rule works on quotients: $\frac{0}{0}$ and $\frac{\infty}{\infty}$. But what about $0 \cdot \infty$ or $\infty - \infty$?
The trick: rewrite them as quotients. A product $f \cdot g$ becomes $\frac{f}{1/g}$. A difference can often be combined into a single fraction. Once you have a quotient, L'Hospital's Rule takes over.
| Property | Value |
|---|---|
| Section | Stewart 6.8 |
| Course | MATH162 |
| Difficulty | Intermediate |
| Time | ~18 minutes |
When you have $\lim_{x \to a} f(x) \cdot g(x)$ where $f(x) \to 0$ and $g(x) \to \infty$:
Method 1: Rewrite as $\frac{f(x)}{1/g(x)}$ to get $\frac{0}{0}$
Method 2: Rewrite as $\frac{g(x)}{1/f(x)}$ to get $\frac{\infty}{\infty}$
Both work! Choose whichever makes the derivatives simpler.
Step 1: Identify the form as $0 \cdot \infty$.
Step 2: Choose which factor to put in the denominator:
Step 3: Apply L'Hospital's Rule to the resulting quotient.
Tip: Pick the conversion that makes differentiation easier!
Problem: Evaluate $\lim_{x \to 0^+} x \ln x$.
Step 1: Identify the 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 $\frac{0}{0}$ but $1/\ln x$ has a complicated derivative
Choose Option A (simpler derivatives): $$x \ln x = \frac{\ln x}{1/x} = \frac{\ln x}{x^{-1}}$$
Step 3: Apply L'Hospital's Rule. $$\lim_{x \to 0^+} \frac{\ln x}{x^{-1}} = \lim_{x \to 0^+} \frac{1/x}{-x^{-2}} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2}$$ $$= \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0$$
Answer: $\lim_{x \to 0^+} x \ln x = 0$
The difference $\infty - \infty$ requires more creativity:
Method 1: Find a common denominator (if the terms are fractions)
Method 2: Multiply by a conjugate (if square roots are involved)
Method 3: Factor out common terms
Step 1: Identify the form as $\infty - \infty$.
Step 2: Look for a way to combine into one fraction:
Step 3: The result is usually $\frac{0}{0}$. Apply L'Hospital's Rule.
Problem: Evaluate $\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{\sin x}\right)$.
Step 1: Identify the form.
Step 2: Combine into one fraction. $$\frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x}$$
Step 3: Check the new form.
Step 4: Apply L'Hospital's Rule. $$\lim_{x \to 0^+} \frac{\sin x - x}{x \sin x} = \lim_{x \to 0^+} \frac{\cos x - 1}{\sin x + x \cos x}$$
Check: $\frac{\cos 0 - 1}{\sin 0 + 0} = \frac{0}{0}$. Apply again: $$= \lim_{x \to 0^+} \frac{-\sin x}{\cos x + \cos x - x \sin x} = \frac{0}{1 + 1 - 0} = 0$$
Answer: $\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{\sin x}\right) = 0$
Problem: Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 2x} - x\right)$.
Step 1: Identify the form.
Step 2: Rationalize by multiplying by conjugate. $$\sqrt{x^2 + 2x} - x = \frac{(\sqrt{x^2 + 2x} - x)(\sqrt{x^2 + 2x} + x)}{\sqrt{x^2 + 2x} + x}$$ $$= \frac{(x^2 + 2x) - x^2}{\sqrt{x^2 + 2x} + x} = \frac{2x}{\sqrt{x^2 + 2x} + x}$$
Step 3: Evaluate directly (or use L'Hospital's if needed).
For $x > 0$: $\sqrt{x^2 + 2x} = x\sqrt{1 + 2/x}$
$$\frac{2x}{x\sqrt{1 + 2/x} + x} = \frac{2x}{x(\sqrt{1 + 2/x} + 1)} = \frac{2}{\sqrt{1 + 2/x} + 1}$$
As $x \to \infty$: $\sqrt{1 + 2/x} \to 1$
$$\lim_{x \to \infty} \frac{2}{\sqrt{1 + 2/x} + 1} = \frac{2}{1 + 1} = 1$$
Answer: $\lim_{x \to \infty} \left(\sqrt{x^2 + 2x} - x\right) = 1$
For $0 \cdot \infty$: Put the factor that's easier to differentiate in the numerator.
| If you have | Try this first |
|---|---|
| $x^n \cdot \ln x$ | $\frac{\ln x}{x^{-n}}$ (power rule for denominator) |
| $x^n \cdot e^{1/x}$ | $\frac{x^n}{e^{-1/x}}$ (chain rule for denominator) |
| Trig times polynomial | Case-by-case (check derivatives) |
For $\infty - \infty$: Look at the structure.
| Structure | Method |
|---|---|
| $\frac{1}{f} - \frac{1}{g}$ | Common denominator |
| $\sqrt{A} - \sqrt{B}$ or $\sqrt{A} - B$ | Conjugate |
| $e^x - P(x)$ | Factor or direct analysis |
Evaluate $\lim_{x \to 0^+} x^2 \ln x$.
Evaluate $\lim_{x \to (\pi/2)^-} (\pi/2 - x) \tan x$.
Evaluate $\lim_{x \to 0} \left(\csc x - \cot x\right)$.
Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 3x} - \sqrt{x^2 - 2x}\right)$.
Evaluate $\lim_{x \to \infty} x^2 e^{-x}$.
Show that converting to $\frac{x^2}{e^x}$ (form $\frac{\infty}{\infty}$) works, but converting to $\frac{e^{-x}}{x^{-2}}$ (form $\frac{0}{0}$) is much harder. Then complete the evaluation.
Question 1: For the limit $\lim_{x \to 0^+} x \ln x$, a student converts it to $\frac{x}{1/\ln x}$. What's wrong with this approach?
The conversion gives form $\frac{0}{1/(-\infty)} = \frac{0}{0^-}$, which is technically $\frac{0}{0}$.
But the derivative of $\frac{1}{\ln x}$ is $\frac{-1}{x(\ln x)^2}$, which is quite messy.
Compare to the standard conversion: $\frac{\ln x}{1/x}$ gives form $\frac{-\infty}{\infty}$. The derivative of $1/x$ is simply $-1/x^2$, which is much cleaner.
Lesson: Both conversions are valid, but one leads to simpler calculations.
Question 2: Why can't we apply L'Hospital's Rule directly to $\lim_{x \to \infty}(x - \ln x)$, even though it's $\infty - \infty$?
L'Hospital's Rule only applies to quotients, not differences. We'd need to first convert this to a quotient form.
One approach: Write as $x(1 - \frac{\ln x}{x})$. Since $\lim_{x \to \infty} \frac{\ln x}{x} = 0$ (by L'Hospital's), we get $x(1 - 0) = x \to \infty$.
Or factor: $\lim_{x \to \infty}(x - \ln x) = \lim_{x \to \infty} x\left(1 - \frac{\ln x}{x}\right) = \infty \cdot 1 = \infty$
The Quotient Gateway: L'Hospital's Rule only works on quotients. Think of $0 \cdot \infty$ and $\infty - \infty$ as being "locked outside." You need to find the right key (conversion) to get them through the gateway into quotient form.
Looking back:
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|---|---|---|
| L'Hospital's Rule | Skills Index | Indeterminate Powers |
Last updated: 2026-01-22