Difficulty: Intermediate | Prerequisites: Part 1 of these notes (L'Hôpital's Rule, 0/0 and ∞/∞ forms), basic differentiation, logarithm properties
This is the second set of notes on Section 4.4. It covers the remaining five indeterminate forms: 0 · ∞, ∞ − ∞, 0⁰, ∞⁰, and 1^∞. All of these are handled by converting them into the 0/0 or ∞/∞ forms from Part 1.
Five more indeterminate forms (0 · ∞, ∞ − ∞, 0⁰, ∞⁰, 1^∞) look unsolvable at first, but each one has a standard trick that converts it into 0/0 or ∞/∞, at which point you use L'Hôpital's Rule from Part 1. Products become quotients by flipping one factor into the denominator. Differences become quotients via common denominators or rationalisation. Powers become products by taking the natural logarithm.
Indeterminate product (type 0 · ∞)
A limit of the form lim f(x) · g(x) where f(x) → 0 and g(x) → ∞ (or −∞). One factor is pulling toward zero while the other pulls toward infinity. Think of it as a tug of war: the result depends on which factor "wins."
Indeterminate difference (type ∞ − ∞)
A limit of the form lim [f(x) − g(x)] where both f(x) → ∞ and g(x) → ∞. Two large quantities are being subtracted, and the result could be anything: ∞ (if f grows faster), −∞ (if g grows faster), or a finite number (if they nearly cancel). In simple terms, subtracting two huge numbers can give any answer.
Indeterminate power of type 0⁰
A limit of the form lim f(x)^g(x) where f(x) → 0⁺ and g(x) → 0. The base wants to make the expression zero, but the exponent wants to make it 1 (since anything to the power 0 is 1). The outcome depends on the rates.
Indeterminate power of type ∞⁰
A limit of the form lim f(x)^g(x) where f(x) → ∞ and g(x) → 0. The base pushes toward infinity, but the exponent pushes toward 1. Another tug of war.
Indeterminate power of type 1^∞
A limit of the form lim f(x)^g(x) where f(x) → 1 and g(x) → ∞. The base is close to 1, which normally keeps things stable, but it is being raised to an ever-larger power, which amplifies even tiny deviations from 1. The classic example is lim(x→∞) (1 + 1/x)ˣ = e.
When you have lim f(x) · g(x) with f(x) → 0 and g(x) → ∞, rewrite the product as a quotient. You have two choices:
Write f · g as f / (1/g), giving a 0/0 form.
Write f · g as g / (1/f), giving an ∞/∞ form.
Both are valid, but one usually produces a simpler derivative than the other. Try the version that avoids a messy chain rule.
Example: lim(x→0⁺) x ln x
This is a 0 · (−∞) form. Rewrite as ln x / (1/x), giving (−∞)/∞. Apply L'Hôpital's Rule: lim(x→0⁺) (1/x) / (−1/x²) = lim(x→0⁺) (−x) = 0.
The alternative rewrite x / (1/ln x) would work eventually but produces a messier derivative.
Example: lim(x→−∞) x eˣ
This is (−∞) · 0 form. Two options: x/e⁻ˣ or eˣ/x⁻¹. The first gives (−∞)/∞, and after L'Hôpital's Rule: lim(x→−∞) 1/e⁻ˣ · (−1) = lim(x→−∞) −eˣ = 0. The second rewrite eˣ/(1/x) leads to 0/0, then differentiating gives eˣ/(−1/x²), which does not simplify cleanly.
When you have lim [f(x) − g(x)] with both f(x) → ∞ and g(x) → ∞, convert the difference into a quotient. Standard techniques:
Common denominator: if f and g are fractions, combine them over a shared denominator.
Rationalisation: if roots are involved, multiply by the conjugate.
Factor out a common term: if there is a shared factor, pull it out.
The goal is to reach a 0/0 or ∞/∞ form so that L'Hôpital's Rule applies.
Example: lim(x→1⁺) (1/ln x − 1/(x − 1))
Both terms → ∞ as x → 1⁺. Combine over a common denominator: [(x − 1) − ln x] / [(ln x)(x − 1)]. As x → 1⁺, the numerator → 0 and the denominator → 0, giving a 0/0 form. Apply L'Hôpital's Rule from there.
Example: lim(x→0⁺) (1/x − 1/(eˣ − 1))
Both terms → ∞. Combine: [(eˣ − 1) − x] / [x(eˣ − 1)]. Both top and bottom → 0, so apply L'Hôpital's Rule. This requires two applications, leading to an answer of 1/2.
All three types use the same strategy:
Let y = f(x)^g(x).
Take the natural log: ln y = g(x) · ln f(x).
The right-hand side is now a product, which is an indeterminate product (0 · ∞ or similar). Evaluate it using the product technique above.
Once you find lim ln y = L, the original limit is y = e^L.
Type ∞⁰ example: lim(x→∞) x^(1/x)
Let y = x^(1/x). Then ln y = (1/x) ln x = (ln x)/x. This is ∞/∞. By L'Hôpital's Rule: lim(x→∞) (1/x)/1 = 0. So lim ln y = 0, meaning y → e⁰ = 1.
Type 1^∞ example: lim(x→∞) (1 + 1/x)ˣ
Let y = (1 + 1/x)ˣ. Then ln y = x · ln(1 + 1/x). This is ∞ · 0. Rewrite as ln(1 + 1/x) / (1/x), which is 0/0. Apply L'Hôpital's Rule:
lim(x→∞) [1/(1 + 1/x)] · (−1/x²) / (−1/x²) = lim(x→∞) 1/(1 + 1/x) = 1.
So lim ln y = 1, meaning y → e¹ = e. This is the definition of the number e.
Type 0⁰ example: lim(x→0⁺) xˣ
Let y = xˣ. Then ln y = x ln x. This is 0 · (−∞). Rewrite as (ln x)/(1/x), giving (−∞)/∞. L'Hôpital's Rule: lim(x→0⁺) (1/x)/(−1/x²) = lim(x→0⁺) (−x) = 0. So lim ln y = 0, meaning y → e⁰ = 1.
More 1^∞ examples:
lim(x→0⁺) (1 − 8x)^(1/x) = e⁻⁸. Set ln y = (1/x) ln(1 − 8x), rewrite as 0/0, apply the rule.
lim(y→0⁺) (cos 2y)^(1/y²) = e⁻². Same strategy with ln y = (1/y²) ln(cos 2y).
lim(y→3⁺) (y − 2)^(1/(y² − 9)) = e^(1/6).
Form | Method |
|---|---|
0/0 | Apply L'Hôpital's Rule directly |
∞/∞ | Apply L'Hôpital's Rule directly |
0 · ∞ | Rewrite as a quotient to get 0/0 or ∞/∞ |
∞ − ∞ | Convert to a quotient via common denominator, rationalisation, or factoring |
0⁰ | Take ln to get an indeterminate product, then exponentiate |
∞⁰ | Take ln to get an indeterminate product, then exponentiate |
1^∞ | Take ln to get an indeterminate product, then exponentiate |
Students often try to apply L'Hôpital's Rule directly to a 0 · ∞ product or an ∞ − ∞ difference without converting to a quotient first. The rule only works on 0/0 or ∞/∞ quotients.
When rewriting a product f · g as a quotient, students sometimes pick the rewrite that makes the derivative harder. Both f/(1/g) and g/(1/f) are correct, but one is usually much cleaner. If your first choice leads to a mess, try the other.
For indeterminate powers, students forget the final step: exponentiating. After finding lim ln y = L, the answer is e^L, not L itself.
Students sometimes assume that 0⁰ = 1 or that 1^∞ = 1 because "anything to the zero is 1" or "1 to anything is 1." These are indeterminate forms precisely because those intuitions break down in the limit. The answer depends on the specific functions involved.
The limit lim(x→∞) (1 + 1/x)ˣ = e is a course classic and appears in many variations. Be ready for limits of the form (1 + k/x)ˣ or (1 − ax)^(1/x).
Expect at least one problem where you must identify the indeterminate form and choose the right conversion strategy. The exam tests whether you know the method, not just the mechanics.
The "take ln, then exponentiate" technique for indeterminate powers is heavily tested. Practise it until the steps are automatic.
Problems asking lim(x→0⁺) x^x are popular because they combine the 0⁰ form with the ln-product technique and the answer (1) surprises many students.
The summary table of all seven forms and their methods is worth memorising. Exams often include a mix of types.
True or False: A limit of the form 0 · ∞ can be evaluated directly with L'Hôpital's Rule.
Answer: False. You must first rewrite the product as a quotient (0/0 or ∞/∞).
Fill in the blank: To handle an indeterminate power, the first step is to let y = f(x)^g(x) and take the ______ of both sides.
Answer: natural logarithm (ln).
True or False: 1^∞ always equals 1.
Answer: False. It is indeterminate. For example, lim(x→∞) (1 + 1/x)ˣ = e, not 1.
Fill in the blank: After finding that lim ln y = 3, the original limit is y = ______.
Answer: e³.
True or False: For a 0 · ∞ product f · g, rewriting as f/(1/g) and g/(1/f) always give the same final answer.
Answer: True (both are algebraically equivalent), though one rewrite may be far easier to differentiate.
Q: Evaluate lim(x→0⁺) x ln x.
A: Form 0 · (−∞). Rewrite as (ln x)/(1/x), giving (−∞)/∞. L'Hôpital's Rule: lim(x→0⁺) (1/x)/(−1/x²) = lim(x→0⁺) (−x) = 0.
Q: Evaluate lim(x→−∞) x eˣ.
A: Form (−∞) · 0. Rewrite as x/e⁻ˣ, giving (−∞)/∞. L'Hôpital's Rule: lim(x→−∞) 1/(−e⁻ˣ) = lim(x→−∞) (−eˣ) = 0.
Q: Evaluate lim(x→0⁺) (1/x − 1/(eˣ − 1)).
A: Form ∞ − ∞. Common denominator: [(eˣ − 1) − x] / [x(eˣ − 1)]. Form 0/0. Two applications of L'Hôpital's Rule give 1/2.
Q: Evaluate lim(x→∞) x^(1/x).
A: Form ∞⁰. Let y = x^(1/x), so ln y = (ln x)/x. Form ∞/∞. L'Hôpital's Rule: lim (1/x)/1 = 0. So y → e⁰ = 1.
Q: Evaluate lim(x→∞) (1 + 1/x)ˣ.
A: Form 1^∞. Let y = (1 + 1/x)ˣ, so ln y = x ln(1 + 1/x). Rewrite as ln(1 + 1/x)/(1/x), form 0/0. L'Hôpital's Rule gives lim 1/(1 + 1/x) = 1. So y → e¹ = e.
Q: Evaluate lim(x→0⁺) xˣ.
A: Form 0⁰. Let y = xˣ, so ln y = x ln x. This is 0 · (−∞). Rewrite as (ln x)/(1/x), form (−∞)/∞. L'Hôpital's Rule: lim (1/x)/(−1/x²) = lim (−x) = 0. So y → e⁰ = 1.
Q: Evaluate lim(x→0⁺) (1 − 8x)^(1/x).
A: Form 1^∞. Let y = (1 − 8x)^(1/x), so ln y = ln(1 − 8x)/x. Form 0/0. L'Hôpital's Rule: lim(x→0⁺) [−8/(1 − 8x)]/1 = −8. So y → e⁻⁸.
Q: Evaluate lim(x→∞) (eˣ + x)^(8/x).
A: Form ∞⁰. Let y = (eˣ + x)^(8/x), so ln y = (8/x) ln(eˣ + x). For large x, eˣ + x ≈ eˣ, so ln(eˣ + x) ≈ x. Then ln y ≈ 8x/x = 8. More rigorously, apply L'Hôpital's Rule to ln(eˣ + x)/x (form ∞/∞) to get lim(eˣ + 1)/(eˣ + x) = 1, so lim ln y = 8. Therefore y → e⁸.
The 1^∞ form connects directly to the definition of e = lim(x→∞) (1 + 1/x)ˣ, which reappears in compound interest (finance), population growth (biology), and radioactive decay (physics). Variations of this limit show up throughout the course.
The "take ln" technique for indeterminate powers is also the foundation of logarithmic differentiation, which you will use when differentiating functions like x^x or (sin x)^(cos x) in later sections.
The summary table of all seven indeterminate forms is the complete toolkit for evaluating difficult limits. Every limit problem in this section reduces, eventually, to one of the two basic forms (0/0 or ∞/∞) that L'Hôpital's Rule handles directly.
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