Indeterminate Differences (∞-∞)

5 min read

When Two Infinities Collide

What happens when you subtract infinity from infinity? Your first instinct might be “zero”; they should cancel out, right?

Not so fast. Consider these two limits:

$$\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{x^2}\right) \quad \text{vs.} \quad \lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{2x}\right)$$

Both are $\infty - \infty$. But the first equals $-\infty$ (the $1/x^2$ term dominates), while the second equals $+\infty$ (they don’t cancel; there’s leftover). The form $\infty - \infty$ is genuinely indeterminate because the answer depends on how fast each term grows.

The solution: convert the difference into a quotient, then apply L’Hospital’s Rule.

Prerequisite Map

Quick Reference

Key Concepts

Why $\infty - \infty$ Is Indeterminate

When $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, the limit $\lim_{x \to a} [f(x) - g(x)]$ could be:

• $+\infty$ (if $f$ grows faster)
• $-\infty$ (if $g$ grows faster)
• A finite number (if they grow at the same rate with a fixed gap)
• Nonexistent (if their difference oscillates)

The key insight: We cannot determine the answer without analyzing the relative growth rates.

Three Strategies for $\infty - \infty$

Strategy 1: Common Denominator

When both terms are fractions, combine them:

$$\frac{1}{f} - \frac{1}{g} = \frac{g - f}{fg}$$

This converts $\infty - \infty$ into a quotient (often $\frac{0}{0}$).

Strategy 2: Rationalization

When square roots are involved, multiply by the conjugate:

$$\sqrt{A} - \sqrt{B} = \frac{(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B})}{\sqrt{A} + \sqrt{B}} = \frac{A - B}{\sqrt{A} + \sqrt{B}}$$

Strategy 3: Factor the Dominant Term

For expressions like $e^x - x$, factor out the dominant piece:

$$e^x - x = e^x\left(1 - \frac{x}{e^x}\right)$$

Then analyze $\frac{x}{e^x}$ separately using L’Hospital’s Rule.

Decision Flowchart

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

Worked Examples

Example 1: Common Denominator Method

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

Step 1: Identify the form.

• $\sec x = \frac{1}{\cos x} \to +\infty$ as $x \to (\pi/2)^-$
• $\tan x = \frac{\sin x}{\cos x} \to +\infty$ as $x \to (\pi/2)^-$
• Form: $\infty - \infty$ ✓

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

Step 3: Verify 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 2: Rationalization Method

Problem: Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 2x} - x\right)$

Step 1: Identify the form.

• $\sqrt{x^2 + 2x} \to \infty$ as $x \to \infty$
• $x \to \infty$
• Form: $\infty - \infty$ ✓

Step 2: Multiply 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: Simplify for large $x > 0$.

Factor $x$ from the square root: $\sqrt{x^2 + 2x} = x\sqrt{1 + 2/x}$ (valid for $x > 0$)

$$= \frac{2x}{x\sqrt{1 + 2/x} + x} = \frac{2x}{x(\sqrt{1 + 2/x} + 1)} = \frac{2}{\sqrt{1 + 2/x} + 1}$$

Step 4: Evaluate the limit. $$\lim_{x \to \infty} \frac{2}{\sqrt{1 + 2/x} + 1} = \frac{2}{\sqrt{1 + 0} + 1} = \frac{2}{2} = 1$$

Answer: $\lim_{x \to \infty} \left(\sqrt{x^2 + 2x} - x\right) = 1$

Example 3: Factoring Method

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

Step 1: Identify the form.

• $e^x \to \infty$
• $x \to \infty$
• Form: $\infty - \infty$ ✓

Step 2: Factor out the dominant term.

Since $e^x$ grows much faster than $x$, factor it out: $$e^x - x = e^x\left(1 - \frac{x}{e^x}\right)$$

Step 3: Analyze $\frac{x}{e^x}$ separately.

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

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

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

Common Mistakes

Practice Problems

Conceptual Questions (CCI-Style)

Question 1: A student claims that $\lim_{x \to \infty}(\sqrt{x^2 + 1} - x) = 0$ because “both terms become infinite, so they cancel.” What is the error in this reasoning, and what is the correct answer?

Question 2: Explain why L’Hospital’s Rule cannot be applied directly to $\lim_{x \to \infty}(e^x - x^2)$, and describe the correct strategy.

Common Misconceptions

Mastery Checklist

• I can identify $\infty - \infty$ as an indeterminate form
• I can use common denominators to combine fractional differences
• I can rationalize differences involving square roots using conjugates
• I can factor out dominant terms to analyze exponential differences
• I can verify the form after conversion before applying L’Hospital’s Rule
• I understand that different $\infty - \infty$ limits can have different answers

Mental Model

The “Tug-of-War” Analogy:

Think of $f(x) - g(x)$ where both approach infinity as a tug-of-war between two strong teams. Just because both teams are “infinitely strong” doesn’t tell you who wins; it depends on which team is stronger.

• If $f$ grows faster: $f$ wins, limit is $+\infty$
• If $g$ grows faster: $g$ wins, limit is $-\infty$
• If they grow at the same rate: it’s a tie, and the margin (finite difference) determines the answer

Converting to a quotient is like setting up a fair scoring system to determine the winner.

Connections

Looking back:

• L’Hospital’s Rule: the tool we apply after conversion
• Recognizing Indeterminate Forms: identifying when $\infty - \infty$ occurs

Looking ahead:

• Improper Integrals: often need limit techniques
• Curve Sketching: analyzing asymptotic behavior

Real-world connections:

• In physics: The electric potential of two opposite charges involves $\infty - \infty$ type expressions near the charges
• In economics: Comparing two growth models (revenue vs. cost) at large scale

Last updated: 2026-01-23