Indeterminate Differences (∞-∞)

MATH162

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

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Indeterminate Differences (∞-∞)

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

Property	Value
Section	Stewart 6.8
Course	MATH162
Difficulty	Intermediate
Time	~20 minutes

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 $\infty - \infty$

Structure	Strategy	Example
$\frac{1}{f} - \frac{1}{g}$	Common denominator	$\frac{1}{x} - \frac{1}{\sin x}$
$\sqrt{A} - \sqrt{B}$	Rationalize (conjugate)	$\sqrt{x^2+x} - x$
$e^x - P(x)$ or similar	Factor dominant term	$e^x - x$

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

Mistake	Why It's Wrong	Correct Approach
Assuming $\infty - \infty = 0$	Different "speeds" of growth matter	Always convert to quotient first
Applying L'Hospital directly to difference	L'Hospital only works on quotients	Convert to $\frac{f}{g}$ form first
Forgetting to check form after conversion	The new form might not be indeterminate	Verify you have $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Using wrong conjugate	Must multiply by $\sqrt{A} + \sqrt{B}$, not subtract	Remember: $(a-b)(a+b) = a^2 - b^2$

Practice Problems

Level 1 Basic Common Denominator

Evaluate $\lim_{x \to 0} (\csc x - \cot x)$.

Thought Process

$\csc x = \frac{1}{\sin x} \to \infty$ and $\cot x = \frac{\cos x}{\sin x} \to \infty$ as $x \to 0$.

Form: $\infty - \infty$

Common denominator: $\frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}$

Check: Both numerator and denominator $\to 0$. Form: $\frac{0}{0}$ ✓

Show Answer

Form: $\infty - \infty$ ✓

Combine: $$\csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}$$

Check: Form is $\frac{0}{0}$ ✓

Apply L'Hospital's Rule: $$\lim_{x \to 0} \frac{1 - \cos x}{\sin x} = \lim_{x \to 0} \frac{\sin x}{\cos x} = \frac{0}{1} = \boxed{0}$$

Level 2 Logarithmic Difference

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

Thought Process

As $x \to 1$: $\ln x \to 0$, so $\frac{1}{\ln x} \to \pm\infty$. Also: $x - 1 \to 0$, so $\frac{1}{x-1} \to \pm\infty$.

Form: $\infty - \infty$

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

Check form at $x = 1$: numerator = $0 - 0 = 0$, denominator = $0 \cdot 0 = 0$.

Form: $\frac{0}{0}$ ✓

Show Answer

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's Rule: $$= \lim_{x \to 1} \frac{1 - 1/x}{\ln x + (x-1)/x} = \lim_{x \to 1} \frac{(x-1)/x}{\ln x + (x-1)/x}$$

Still $\frac{0}{0}$. Apply again:

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

Denominator derivative: $\frac{1}{x} + \frac{1}{x^2}$

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

Level 3 Rationalization with Coefficients

Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + 5x} - \sqrt{x^2 - 3x}\right)$.

Thought Process

Both square roots $\to \infty$ as $x \to \infty$. Form: $\infty - \infty$

Rationalize: multiply by $\frac{\sqrt{x^2+5x} + \sqrt{x^2-3x}}{\sqrt{x^2+5x} + \sqrt{x^2-3x}}$

Numerator becomes: $(x^2 + 5x) - (x^2 - 3x) = 8x$

For large $x$: factor $x$ from each square root.

Show Answer

Form: $\infty - \infty$ ✓

Rationalize: $$= \frac{(x^2 + 5x) - (x^2 - 3x)}{\sqrt{x^2+5x} + \sqrt{x^2-3x}} = \frac{8x}{\sqrt{x^2+5x} + \sqrt{x^2-3x}}$$

Factor $x$ from square roots (for $x > 0$): $$= \frac{8x}{x\sqrt{1+5/x} + x\sqrt{1-3/x}} = \frac{8}{\sqrt{1+5/x} + \sqrt{1-3/x}}$$

Evaluate: $$\lim_{x \to \infty} \frac{8}{\sqrt{1+0} + \sqrt{1-0}} = \frac{8}{1 + 1} = \boxed{4}$$

Level 4 Mixed Transcendentals

Evaluate $\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1}\right)$.

Thought Process

As $x \to 0^+$:

$\frac{1}{x} \to +\infty$
$e^x - 1 \to 0^+$, so $\frac{1}{e^x - 1} \to +\infty$

Form: $\infty - \infty$

Common denominator: $\frac{e^x - 1 - x}{x(e^x - 1)}$

Check: numerator $\to 0$ (using Taylor: $e^x \approx 1 + x + x^2/2$) denominator $\to 0$

Show Answer

Form: $\infty - \infty$ ✓

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

Check form: Both numerator ($e^0 - 1 - 0 = 0$) and denominator ($0 \cdot 0 = 0$) approach 0.

Apply L'Hospital's Rule: $$= \lim_{x \to 0^+} \frac{e^x - 1}{(e^x - 1) + xe^x} = \lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x}$$

At $x = 0$: $\frac{0}{0 + 0} = \frac{0}{0}$. Apply again:

$$= \lim_{x \to 0^+} \frac{e^x}{e^x + e^x + xe^x} = \lim_{x \to 0^+} \frac{e^x}{2e^x + xe^x}$$

$$= \lim_{x \to 0^+} \frac{1}{2 + x} = \frac{1}{2} = \boxed{\frac{1}{2}}$$

Level 5 Proof: Growth Rate Comparison

Prove that $\lim_{x \to \infty} \left[x - x^2\ln\left(1 + \frac{1}{x}\right)\right] = \frac{1}{2}$.

Hint: Use the Taylor expansion $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$ for small $u$.

Thought Process

Let $u = 1/x$. As $x \to \infty$, $u \to 0$.

$\ln(1 + 1/x) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \cdots$

$x^2 \ln(1 + 1/x) = x - \frac{1}{2} + \frac{1}{3x} - \cdots$

$x - x^2\ln(1 + 1/x) = x - \left(x - \frac{1}{2} + \frac{1}{3x} - \cdots\right) = \frac{1}{2} - \frac{1}{3x} + \cdots$

Show Answer

Form: As $x \to \infty$: $x \to \infty$ and $x^2\ln(1 + 1/x) \to \infty$ (verify: $\ln(1 + 1/x) \approx 1/x$, so $x^2 \cdot (1/x) = x$). Form: $\infty - \infty$ ✓

Method 1: Taylor Series

For small $u$: $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots$

Let $u = 1/x$ (which is small as $x \to \infty$):

$$\ln\left(1 + \frac{1}{x}\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} - \cdots$$

Multiply by $x^2$:

$$x^2\ln\left(1 + \frac{1}{x}\right) = x - \frac{1}{2} + \frac{1}{3x} - \frac{1}{4x^2} + \cdots$$

Subtract from $x$:

$$x - x^2\ln\left(1 + \frac{1}{x}\right) = x - \left(x - \frac{1}{2} + \frac{1}{3x} - \cdots\right) = \frac{1}{2} - \frac{1}{3x} + \cdots$$

As $x \to \infty$: $\boxed{\frac{1}{2}}$

Method 2: L'Hospital's Rule

Rewrite as a quotient: Let $t = 1/x$, so as $x \to \infty$, $t \to 0^+$.

$$x - x^2\ln(1 + 1/x) = \frac{1}{t} - \frac{1}{t^2}\ln(1 + t) = \frac{t - \ln(1+t)}{t^2}$$

Form: $\frac{0 - 0}{0} = \frac{0}{0}$ ✓

Apply L'Hospital twice to get $\frac{1}{2}$. $\square$

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?

Answer

The error is assuming all infinities are "equal." The correct calculation:

$$\sqrt{x^2 + 1} - x = \frac{1}{\sqrt{x^2+1} + x} \to \frac{1}{\infty} = 0$$

So the answer IS 0, but the reasoning was wrong. The correct reason: after rationalization, the numerator is 1 (constant), while the denominator grows without bound.

A slight change like $\sqrt{x^2 + x} - x$ would give $\frac{1}{2}$, not 0. The coefficient matters!

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.

Answer

L'Hospital's Rule applies to quotients, not differences. The expression $e^x - x^2$ is a difference, so we cannot differentiate numerator/denominator.

Correct strategy: Factor out the dominant term.

$$e^x - x^2 = e^x\left(1 - \frac{x^2}{e^x}\right)$$

Now analyze $\frac{x^2}{e^x}$ using L'Hospital (form $\frac{\infty}{\infty}$):

$$\lim_{x \to \infty}\frac{x^2}{e^x} = \lim_{x \to \infty}\frac{2x}{e^x} = \lim_{x \to \infty}\frac{2}{e^x} = 0$$

So $e^x - x^2 = e^x(1 - 0) \to \infty$.

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

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Last updated: 2026-01-23