Difficulty: Intermediate to Advanced | Prerequisites: Part 1 of these notes (core L'Hôpital's Rule for 0/0 and ∞/∞), logarithmic differentiation, algebraic manipulation of fractions.
L'Hôpital's Rule handles 0/0 and ∞/∞ directly, but calculus exams routinely throw limits that start as other indeterminate forms: 0 · ∞, ∞ - ∞, 0⁰, ∞⁰, and 1^∞. None of these can go straight into L'Hôpital's Rule. The trick in every case is algebraic or logarithmic manipulation that converts the expression into a 0/0 or ∞/∞ quotient, at which point the core rule takes over. If you can handle Part 1 comfortably, this is the next layer of the same idea.
Five more indeterminate forms (0 · ∞, ∞ - ∞, 0⁰, ∞⁰, 1^∞) cannot be fed directly into L'Hôpital's Rule. Each has a standard conversion: products become quotients by rewriting f · g as f/(1/g), differences become quotients via common denominators or factoring, and powers become products by taking the natural logarithm. Once converted, the core rule finishes the job.
Indeterminate product (0 · ∞)
Occurs when one factor approaches 0 and the other approaches ±∞ as x approaches some value. The product could be anything, so the form is indeterminate. In simple terms, something shrinking times something growing: the outcome depends on who wins the race.
Indeterminate difference (∞ - ∞)
Occurs when two terms both approach ∞ (or both approach -∞) and you are subtracting them. The difference could be zero, finite, or infinite. Think of it as: two quantities both blowing up, but you are looking at the gap between them, which is not obvious.
Indeterminate power (0⁰, ∞⁰, 1^∞)
Occurs when an expression of the form f(x)^g(x) has a base and exponent that together produce one of these three forms. All three are indeterminate because the base and exponent pull the result in competing directions. In simple terms, the base wants one thing, the exponent wants another, and the limit could land anywhere.
Logarithmic trick for indeterminate powers
The technique of setting y = lim f(x)^g(x), taking ln of both sides to get ln y = lim g(x) · ln f(x), solving the resulting 0 · ∞ product, and then exponentiating the answer. Think of it as: the natural log turns a power problem into a multiplication problem, which you already know how to handle.
When lim f(x) · g(x) gives a 0 · ∞ form, rewrite the product as a quotient:
f · g = f / (1/g), which gives a 0/0 form, or
f · g = g / (1/f), which gives an ∞/∞ form
Choose whichever makes the derivative easier to compute.
Worked example: lim x→∞ x · tan(4/x)
Direct evaluation: ∞ · 0. Indeterminate product.
Rewrite: x · tan(4/x) = tan(4/x) / (1/x). Now as x → ∞, both top and bottom → 0. This is 0/0.
Differentiate: numerator derivative = sec²(4/x) · (-4/x²). Denominator derivative = -1/x².
Simplify: the -1/x² terms cancel, leaving lim x→∞ 4sec²(4/x).
As x → ∞, 4/x → 0, so sec²(0) = 1. Answer: 4.
When lim [f(x) - g(x)] gives an ∞ - ∞ form, convert the difference into a quotient using one of three methods:
Factor out a common factor
Find a common denominator
Rationalise (multiply by a conjugate)
Worked example A: lim x→∞ (xe^(6/x) - x)
Direct evaluation: ∞ - ∞. Indeterminate difference.
Factor: x(e^(6/x) - 1). Now this is ∞ · 0, an indeterminate product.
Rewrite as quotient: (e^(6/x) - 1) / (1/x). As x → ∞, both top and bottom → 0. This is 0/0.
Differentiate: numerator = e^(6/x) · (-6/x²). Denominator = -1/x².
The -1/x² cancels, leaving lim x→∞ 6e^(6/x) = 6 · e⁰ = 6. Answer: 6.
Worked example B: lim x→1⁺ [x/(x-1) - 1/ln(x)]
Direct evaluation: ∞ - ∞. Indeterminate difference.
Common denominator: (x · ln x - (x - 1)) / ((x - 1) · ln x). As x → 1⁺, both top and bottom → 0. This is 0/0.
Apply L'Hôpital's Rule. After differentiation and simplification, the limit reduces to ln(x) / (ln x + 1 - 1/x), still 0/0.
Apply the rule again. The final result is 1/2. Answer: 1/2.
When lim f(x)^g(x) gives one of the three indeterminate power forms, use the logarithmic method:
Set y = lim f(x)^g(x).
Take ln of both sides: ln y = lim g(x) · ln f(x).
The right side is now a 0 · ∞ indeterminate product. Convert it to a quotient and apply L'Hôpital's Rule to find some value L.
Since ln y = L, then y = e^L. That is your answer.
Worked example A (1^∞ form): lim x→0 (1 - 6x)^(1/x)
Direct evaluation: 1^∞. Indeterminate power.
Let y = lim x→0 (1 - 6x)^(1/x). Then ln y = lim x→0 (1/x) · ln(1 - 6x).
Rewrite: ln y = lim x→0 ln(1 - 6x) / x. This is 0/0.
Differentiate: numerator = (-6)/(1 - 6x). Denominator = 1.
ln y = lim x→0 (-6)/(1 - 6x) = -6/(1) = -6.
Therefore y = e^(-6). Answer: e^(-6).
Worked example B (1^∞ form, general): lim x→∞ (1 + a/x)^(bx), where a and b are real numbers
Direct evaluation: 1^∞. Indeterminate power.
Let y = lim x→∞ (1 + a/x)^(bx). Then ln y = lim x→∞ bx · ln(1 + a/x). This is ∞ · 0.
Rewrite: ln y = lim x→∞ b · ln(1 + a/x) / (1/x). This is 0/0.
Differentiate: numerator = b · [1/(1 + a/x)] · (-a/x²). Denominator = -1/x².
The -1/x² cancels. ln y = lim x→∞ ab / (1 + a/x) = ab / 1 = ab.
Therefore y = e^(ab). Answer: e^(ab).
This is the general result: lim x→∞ (1 + a/x)^(bx) = e^(ab). When a = 1 and b = 1, this gives the classic definition of e.
Converting indeterminate products to quotients:
f \cdot g = \frac{f}{1/g} \quad \text{or} \quad f \cdot g = \frac{g}{1/f}Logarithmic method for indeterminate powers:
y = \lim_{x \to a} f(x)^{g(x)} \implies \ln y = \lim_{x \to a} g(x) \cdot \ln f(x)Once you find ln y = L, the answer is y = e^L.
General result for 1^∞ compound interest form:
\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^{bx} = e^{ab}The 1^∞ form appears in compound interest. The formula lim n→∞ (1 + r/n)^(nt) = e^(rt) is the continuous compounding formula used throughout finance, and it is a direct application of the general result proved with L'Hôpital's Rule. The 0 · ∞ form shows up in probability theory (moments of distributions that decay exponentially but are weighted by polynomial terms) and in physics (damped oscillations where amplitude decays while frequency grows).
Students often assume that 0 · ∞ = 0 or that ∞ - ∞ = 0. Neither is true in general. These are indeterminate forms, meaning the answer depends on the specific functions involved.
When converting a product f · g into a quotient, students sometimes pick the version that makes the derivative harder rather than easier. Try both f/(1/g) and g/(1/f) and see which one simplifies better before committing.
With indeterminate powers, students frequently forget the final step: exponentiating the result. The logarithmic method gives you ln y = L, not y = L. The answer is y = e^L.
Students sometimes try to apply L'Hôpital's Rule directly to a product or difference without converting to a quotient first. The rule only works on ratios in 0/0 or ∞/∞ form.
⚠️ Exams regularly include at least one "other indeterminate form" problem. The 1^∞ form and the 0 · ∞ form are the most frequently tested.
⚠️ The logarithmic trick for indeterminate powers is a multi-step process. Exams will test whether you can execute all four steps (set y, take ln, solve with L'Hôpital, exponentiate) without losing track.
⚠️ Converting ∞ - ∞ into a quotient via common denominator is algebraically heavy. Practice the algebra so it does not slow you down under time pressure.
⚠️ Know the general result lim x→∞ (1 + a/x)^(bx) = e^(ab). It appears as a standalone problem and as a step inside larger problems.
True or False: 0 · ∞ is always equal to 0. (False, it is an indeterminate form.)
Fill in the blank: To apply L'Hôpital's Rule to a 0 · ∞ product, you must first rewrite it as a ______. (quotient, in 0/0 or ∞/∞ form)
True or False: When using the logarithmic method for indeterminate powers, if ln y = -3, then y = -3. (False, y = e^(-3).)
Fill in the blank: lim x→∞ (1 + 5/x)^(2x) = ______. (e^10)
True or False: ∞ - ∞ = 0. (False, it is indeterminate.)
Q: Evaluate lim x→0⁺ x · ln(x).
A: This is 0 · (-∞). Rewrite as ln(x) / (1/x), which is -∞/∞. Differentiate: (1/x) / (-1/x²) = -x. Then lim x→0⁺ (-x) = 0.
Q: Evaluate lim x→∞ (x² - 3x). Is this an indeterminate difference?
A: No. As x → ∞, x² - 3x = x(x - 3) → ∞. This is ∞, not ∞ - ∞, because both terms grow and the subtraction does not produce competing infinities. The limit is ∞.
Q: Evaluate lim x→0⁺ x^x.
A: This is the 0⁰ form. Let y = x^x, so ln y = x · ln x. This is 0 · (-∞). Rewrite: ln(x) / (1/x) = -∞/∞. Differentiate: (1/x) / (-1/x²) = -x → 0. So ln y = 0, and y = e⁰ = 1.
Q: Evaluate lim x→∞ (1 + 3/x)^(4x).
A: This is the 1^∞ form. Using the general result with a = 3 and b = 4: the answer is e^(12).
Q: Evaluate lim x→0 [1/x - 1/sin(x)].
A: This is ∞ - ∞. Common denominator: (sin x - x) / (x · sin x). As x → 0, this is 0/0. Apply L'Hôpital's Rule: (cos x - 1) / (sin x + x · cos x). At x = 0 this is still 0/0. Apply again: (-sin x) / (cos x + cos x - x · sin x) = 0 / 2 = 0.
This material connects directly to Part 1 of these notes, since every problem here eventually reduces to a 0/0 or ∞/∞ quotient. The logarithmic method for indeterminate powers uses the same log properties covered in the exponential and logarithmic functions unit. The general result lim (1 + a/x)^(bx) = e^(ab) connects forward to sequences and series, where it appears in convergence tests, and to continuous compounding in applied mathematics and finance.
Indeterminate products, indeterminate differences, indeterminate powers, 0 times infinity, infinity minus infinity, 0 to the 0, infinity to the 0, 1 to the infinity, logarithmic trick for limits, ln method limits, continuous compounding derivation, L'Hôpital other forms, MATH 103, Section 7.8, Penn State Calculus 1, Rimmer, convert product to quotient, common denominator limits