When a limit produces 0/0 or ∞/∞, your calculator gives up: but calculus doesn't. There are exactly seven indeterminate forms, and a different algebraic weapon for each.
All 7 forms12 worked examplesAlgebraic & ln techniques
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§ 01What Makes a Form "Indeterminate"?
An indeterminate form is not an error: it is a signal that the limit could be anything, and that you need more information to decide which.
When you compute the limit of a fraction and both the numerator and denominator approach zero simultaneously, you might be tempted to conclude that the limit is 0/0 = 0, or perhaps that it is undefined. Both conclusions are wrong. The expression 0/0 does not have a fixed numerical value, it is a competition between two quantities both racing toward zero, and the winner depends entirely on how fast each one runs. Different functions reaching 0/0 can give limits of 0, 7, −3, ∞, or any real number whatsoever.
This is what it means for a form to be indeterminate: the symbol alone, without knowing the specific functions involved, tells you nothing about the limiting value. Contrast this with a form like 5/0, which is not indeterminate: it always signals that the limit is ±∞ (or that no limit exists). That form is undefined but not indeterminate.
Three Limits, Same Symbol, Three Different Answers
All three of the following produce the form 0/0 at x = 0, yet give completely different limits:
The lesson is powerful: seeing 0/0 in a limit should trigger a method, not a conclusion. Your job is to transform the expression algebraically until the form resolves into something definite.
§ 02The Seven Indeterminate Forms
Mathematicians have catalogued exactly seven distinct indeterminate forms. Recognising which one you have is the first step in choosing the right resolution technique.
0/0Most common
∞/∞Ratio race
0·∞Product form
∞−∞Difference
0⁰Power form
1∞Power form
∞⁰Power form
Notice that the first two forms (0/0 and ∞/∞) are fractions, they are directly amenable to L'Hôpital's Rule once identified. The next two (0·∞ and ∞−∞) need to be rewritten as fractions before L'Hôpital can be applied. The final three (0⁰, 1∞, ∞⁰) involve exponents and require the logarithm trick: take ln of both sides, compute the new limit, then exponentiate to recover the original limit.
These are NOT indeterminate — do not confuse them
The following expressions have definite, unambiguous values and are not indeterminate:
any non-zero number / 0 → ±∞ · 0/non-zero → 0 ·
∞ + ∞ → ∞ · 1·∞ → ∞ · 0∞ → 0 · ∞∞ → ∞
§ 03Resolving 0/0 — Algebraic Techniques
The 0/0 form is the one you will encounter most frequently. Before reaching for L'Hôpital, always try algebra: it is faster and builds better intuition.
Technique 1 — Factoring and Cancelling
When both numerator and denominator share a common factor that vanishes at the limit point, factoring and cancelling that factor will resolve the 0/0 form immediately. This works for polynomials and rational expressions.
When a square root appears in the numerator or denominator, multiply top and bottom by the conjugate expression. This clears the surd and reveals the cancellable factor.
Example 02Find limx→0 (√(x+4) − 2) / x
At x = 0: (√4 − 2)/0 = 0/0. Rationalise the numerator.
Multiply by conjugate over conjugate
× (√(x+4) + 2) / (√(x+4) + 2)
Numerator becomes difference of squares
(x + 4) − 4 = x
Simplify and cancel x
x / [x · (√(x+4) + 2)] = 1 / (√(x+4) + 2)
Substitute x = 0
1 / (√4 + 2) = 1/4
Limit = 1/4
Example 03Find limx→5 (x² − 3x − 10) / (x − 5)
Substitution gives 0/0. Factor the numerator.
Factor quadratic
x² − 3x − 10 = (x − 5)(x + 2)
Cancel and evaluate
limx→5 (x + 2) = 7
Limit = 7
§ 04Resolving ∞/∞ — Dominance and Division
When both numerator and denominator grow without bound, the limit depends entirely on which one grows faster. There is a clear hierarchy of growth rates in mathematics.
Dominance Hierarchy (slowest → fastest growth)
ln x ≪ xp (any p > 0) ≪ aˣ (a > 1) ≪ x! ≪ xˣ
When the ∞/∞ form involves rational functions (polynomials over polynomials), the quickest technique is to divide every term by the highest power of x in the denominator, then take the limit as x → ∞.
Example 04Find limx→∞ (3x² + 5x − 1) / (2x² − 7)
Both numerator and denominator → ∞. Divide through by x² (highest power in denominator).
Divide numerator and denominator by x²
(3 + 5/x − 1/x²) / (2 − 7/x²)
As x → ∞, all 1/xⁿ terms → 0
(3 + 0 − 0) / (2 − 0) = 3/2
Limit = 3/2
Example 05Find limx→∞ eˣ / x³
Both → ∞. By the dominance hierarchy, eˣ grows far faster than any polynomial. Applying L'Hôpital three times confirms:
Apply L'Hôpital (×3) — differentiating top and bottom each time
eˣ/x³ → eˣ/3x² → eˣ/6x → eˣ/6
Final limit
limx→∞ eˣ/6 = ∞
Limit = ∞ (exponential dominates)
§ 05Resolving 0·∞ and ∞ − ∞
These two forms cannot be fed directly to L'Hôpital. They must first be rewritten as a single fraction (0/0 or ∞/∞), then resolved from there.
The 0·∞ Strategy — Rewrite as a Fraction
If lim f(x) = 0 and lim g(x) = ∞, then lim f(x)·g(x) is 0·∞. Rewrite the product as a fraction by placing one factor in the denominator: either f·g = f / (1/g) giving 0/0, or f·g = g / (1/f) giving ∞/∞. Choose whichever form leads to simpler differentiation.
The ∞ − ∞ Strategy — Combine into a Single Fraction
When two terms both blow up to ∞ but with opposite signs, the difference is indeterminate. Find a common denominator to combine the two terms into a single fraction, which will usually produce a 0/0 or ∞/∞ form.
Example 07Find limx→0 (1/sin x − 1/x)
Both 1/sin x and 1/x → ∞ as x → 0. Form is ∞ − ∞.
Common denominator: x sin x
(x − sin x) / (x sin x) — now 0/0 form
Apply L'Hôpital (numerator: 1 − cos x; denominator: sin x + x cos x)
(1 − cos x) / (sin x + x cos x)
Still 0/0 — apply L'Hôpital again
sin x / (cos x + cos x − x sin x) = sin x / (2cos x − x sin x)
Substitute x = 0
0 / (2 − 0) = 0
Limit = 0
§ 06Power Forms — The Logarithm Trick
The forms 1∞, 0⁰, and ∞⁰ all involve exponents where both base and exponent are competing. The key insight: take the natural log first, then exponentiate at the end.
The Logarithm Trick — Step by Step
Rename the limit. Let L = lim f(x)^g(x). You want to find L.
Take ln of both sides. ln L = lim g(x) · ln f(x). This converts the power form into a 0·∞ form, which is easier.
Resolve the 0·∞ form using the rewrite-as-fraction technique from §05.
Recover L. If you found ln L = k, then L = ek.
Example 08Find limx→0⁺ xx (form 0⁰)
As x → 0⁺: both base x → 0 and exponent x → 0. Form is 0⁰.
Let L = lim xx, take ln
ln L = limx→0⁺ x · ln x = 0 · (−∞)
From Example 06, this limit equals 0
ln L = 0
Exponentiate
L = e⁰ = 1
Limit = 1 (0⁰ → 1 in this context)
Example 09Find limx→∞ (1 + 1/x)x (form 1∞)
This is the famous limit that defines Euler's number e. Base → 1, exponent → ∞. Form is 1∞.
Let L = lim (1 + 1/x)ˣ, take ln
ln L = lim x · ln(1 + 1/x) — form 0·∞
Rewrite: ln(1+1/x) / (1/x) — form 0/0
Apply L'Hôpital: [−1/(x²(1+1/x))] / (−1/x²)
Simplify
= 1 / (1 + 1/x) → 1/(1+0) = 1
So ln L = 1, therefore
L = e¹ = e
Limit = e ≈ 2.71828…
Example 10Find limx→∞ x1/x (form ∞⁰)
Base x → ∞, exponent 1/x → 0. Form is ∞⁰.
ln L = lim (1/x) · ln x = ln x / x — form ∞/∞
Apply L'Hôpital: (1/x) / 1 = 1/x
Limit of 1/x as x → ∞
ln L = 0
Exponentiate
L = e⁰ = 1
Limit = 1
Example 11Find limx→0⁺ (sin x)x (form 0⁰)
ln L = lim x · ln(sin x)
= lim ln(sin x) / (1/x) — form ∞/∞
L'Hôpital: differentiate top and bottom
Numerator: cos x / sin x = cot x · Denominator: −1/x²
Result: cot x · (−x²) = −x² cos x / sin x
As x → 0⁺: −x² cos x / sin x ≈ −x²·1 / x = −x → 0
ln L = 0, so
L = e⁰ = 1
Limit = 1
Example 12Find limx→0 (cos x)1/x² (form 1∞)
As x → 0: cos x → 1 and 1/x² → ∞. Classic 1∞ form.
Mistake 1 — Treating 0/0 as zero
The fraction 0/0 does not simplify to zero. It is indeterminate — you do not know the answer until you work through the algebra or apply L'Hôpital. Students who write "0/0 = 0" are asserting a result they have not earned.
Mistake 2 — Applying L'Hôpital when the form is NOT indeterminate
L'Hôpital's rule only applies to 0/0 or ∞/∞ forms. Applying it to, say, limx→0 (sin x)/2 (which is just 0/2 = 0) would give cos(x)/2, which also equals 1/2 at x = 0 — but you are getting lucky. Apply L'Hôpital to (sin x)/(x) which is genuinely 0/0.
Mistake 3 — Forgetting to exponentiate after the ln trick
The most common error in power-form limits is computing ln L = k and then reporting k as the answer. The actual limit is L = eᵏ, not k. Always exponentiate at the end.
Mistake 4 — Confusing 1∞ with "1 to any power is 1"
The rule "1ⁿ = 1 for all n" applies when the base is exactly 1 and fixed. In a limit, the base approaches 1 but is never exactly 1 — the exponent grows to infinity simultaneously, and the competition between how fast the base leaves 1 and how fast the exponent grows determines the result.
§ 08Practice Quiz — 10 Questions
Mixed multiple-choice and fill-in-the-blank questions covering all seven indeterminate forms and the techniques to resolve them.
Score: 0 / 10
Question 1 of 10 — Multiple Choice
Which of the following is an indeterminate form?
B — 0/0. This is the most fundamental indeterminate form. 5/0 signals ±∞ (not indeterminate). 0/5 = 0 (definite). ∞+∞ = ∞ (definite, not indeterminate).
Question 2 of 10 — Fill in the Blank
limx→2 (x² − 4) / (x − 2) = ?
Answer:
Factor: x² − 4 = (x−2)(x+2). Cancel (x−2), then substitute.
x² − 4 = (x−2)(x+2), so the limit is lim (x+2) = 4.
Question 3 of 10 — Multiple Choice
limx→∞ (5x³ − x) / (2x³ + 9) = ?
B — 5/2. Divide numerator and denominator by x³ (highest power): (5 − 1/x²)/(2 + 9/x³) → 5/2 as x → ∞.
Question 4 of 10 — Multiple Choice
To resolve the 0·∞ form, you should:
B. L'Hôpital only works on 0/0 or ∞/∞ fractions. So f·g = f/(1/g) or g/(1/f) — whichever produces the simpler fraction to differentiate.
Question 5 of 10 — Fill in the Blank
limx→0⁺ x · ln x = ?
Answer:
Rewrite as ln x / (1/x), then apply L'Hôpital. See Example 06 above.
Rewrite: ln x / (1/x) = (1/x) / (−1/x²) = −x → 0. The limit is 0.
Question 6 of 10 — Multiple Choice
Which indeterminate form does limx→∞ (1 + 2/x)ˣ produce?
B — 1∞. As x → ∞: the base (1 + 2/x) → 1 and the exponent x → ∞. This is the 1∞ power form — use the ln trick. (The limit itself equals e².)
Question 7 of 10 — Multiple Choice
After applying the ln trick and finding ln L = 3, the original limit L equals:
C — e³. The ln trick sets L = lim f(x)^g(x), so ln L = 3. To recover L, exponentiate both sides: L = e³. Forgetting this final step is the most common error with power forms.
Question 8 of 10 — Fill in the Blank
limx→∞ (4x² + 1) / (x² − 3x) = ?
Answer:
Divide numerator and denominator by x² (the highest power).
(4 + 1/x²) / (1 − 3/x) → 4/1 = 4.
Question 9 of 10 — Multiple Choice
limx→0⁺ xx equals:
C — 1. See Example 08 above. ln L = lim x·ln x = 0, so L = e⁰ = 1. The 0⁰ form resolved to 1 here — but remember, this is not always the case; it depends entirely on the specific functions.
Question 10 of 10 — Multiple Choice
How many indeterminate forms are there in total?
C — 7. The seven are: 0/0, ∞/∞, 0·∞, ∞−∞, 0⁰, 1∞, and ∞⁰. Each requires a different technique: factoring/rationalising for 0/0, dominance/division for ∞/∞, fraction-rewrite for 0·∞ and ∞−∞, and the ln trick for all three power forms.
Next StepsContinue the Limits Series
Indeterminate forms appear in nearly every advanced limit problem. The pages that follow build directly on this foundation:
L'Hôpital's Rule, the dedicated theorem for resolving 0/0 and ∞/∞ by differentiating top and bottom.
Continuity, understanding when direct substitution is valid in the first place.
Epsilon-Delta Definition, the rigorous foundation behind every limit technique you have learned.