Indeterminate Forms
1. What an Indeterminate Form Is
An indeterminate form is an expression that arises when a limit is evaluated by direct substitution, and whose value cannot be decided from the form itself — the answer depends on how the two competing functions behave. For example, x/x → 1, x²/x → 0 and x/x² → ∞ as x → 0, yet all three look like 0/0.
2. The Seven Indeterminate Forms
| Form | Example (x → 0 or x → ∞) | Value is not determined by the form |
|---|---|---|
| 0/0 | x/x, sin x/x, (1 − cos x)/x² | 1, 1, 1/2 — all different |
| ∞/∞ | (2x)/(x), (x)/(x²), (x² + x)/(3x²) | 2, 0, 1/3 |
| 0 × ∞ | x log x, x · (1/x) | 0, 1 |
| ∞ − ∞ | (1/x − 1/x²), √(x² + x) − x | ∞, 1/2 |
| 1^∞ | (1 + x)^(1/x) | e |
| 0⁰ | x^x | 1 |
| ∞⁰ | x^(1/x) | 1 |
Not indeterminate (these have definite behaviour): 0/∞ = 0, ∞/0 = ∞, ∞ + ∞ = ∞, 0⁰ … careful — 0⁰ is indeterminate, but 0^∞ = 0 and ∞^∞ = ∞ are determinate.
3. Strategy — Reduce Everything to 0/0 or ∞/∞
Only the two basic forms 0/0 and ∞/∞ can be attacked directly by L'Hôpital's rule, so the other five are converted first:
| Form | Conversion |
|---|---|
| 0 × ∞ | write f·g as f/(1/g) → gives 0/0, or as g/(1/f) → gives ∞/∞. Choose whichever differentiates more easily. |
| ∞ − ∞ | combine into a single fraction (common denominator), or rationalise (multiply by the conjugate), or factor out the dominant term |
| 1^∞, 0⁰, ∞⁰ | put y = f(x)^g(x), take logarithms: log y = g(x)·log f(x) which is a 0 × ∞ form; find L = lim log y, then the required limit is e^L |
4. Other Tools (often faster than L'Hôpital)
- Factorisation and cancellation: (x² − 4)/(x − 2) = x + 2 for x ≠ 2.
- Rationalisation: multiply by the conjugate — the standard trick for √(x² + x) − x.
- Standard limits (memorise these):
| Limit (x → 0) | Value |
|---|---|
| sin x / x, tan x / x, sin⁻¹x / x, tan⁻¹x / x | 1 |
| (1 − cos x)/x² | 1/2 |
| (e^x − 1)/x, log(1 + x)/x | 1 |
| (a^x − 1)/x | log a |
| ((1 + x)ⁿ − 1)/x | n (any real n) |
| (1 + x)^(1/x), x → 0 | e |
| (1 + 1/x)^x, x → ∞ | e |
| (1 + a/x)^x, x → ∞ | e^a |
- Series expansion (Maclaurin): replace sin x, cos x, e^x, log(1+x), (1+x)ⁿ by their expansions and cancel powers of x — often the quickest route for 0/0.
5. Worked Examples — One for Each Form
(a) 0/0 : lim x→2 (x² - 4)/(x - 2) = lim (x - 2)(x + 2)/(x - 2) = lim (x + 2) = 4
[factorise rather than differentiate]
(b) ∞/∞ : lim x→∞ (3x² + 2x)/(5x² - x + 1)
divide numerator and denominator by x²:
= lim (3 + 2/x)/(5 - 1/x + 1/x²) = 3/5
["degree rule": for equal degrees, the limit is the ratio of leading coefficients]
(c) 0 × ∞ : lim x→0+ x log x = lim (log x)/(1/x) [now ∞/∞]
= lim (1/x)/(-1/x²) = lim (-x) = 0 ✓
(d) ∞ − ∞ : lim x→∞ [√(x² + x) - x]
rationalise: = lim [ (x² + x) - x² ] / [√(x² + x) + x]
= lim x / [√(x² + x) + x]
divide by x: = lim 1 / [√(1 + 1/x) + 1] = 1/2 ✓
(e) 1^∞ : lim x→0 (1 + x)^(1/x) = e [standard limit; via logs:
log y = (1/x) log(1 + x) → 1, so y → e¹ = e]
(f) 0⁰ : lim x→0+ x^x : let y = x^x ⇒ log y = x log x → 0 (example (c))
⇒ y → e⁰ = 1 ✓
(g) ∞⁰ : lim x→∞ x^(1/x) : log y = (log x)/x → 0 by L'Hôpital (1/x)/1 = 0
⇒ y → e⁰ = 1 ✓
6. Reported Answers to Remember
| Limit | Result |
|---|---|
| lim x→0 (sin x)/x | 1 |
| lim x→0 (1 − cos x)/x² | 1/2 |
| lim x→∞ (3x² + 2x)/(5x² − x + 1) | 3/5 |
| lim x→0⁺ x log x, lim x→0⁺ x^x | 0 and 1 |
| lim x→∞ (√(x² + x) − x) | 1/2 |
| lim x→0 (1 + x)^(1/x), lim x→∞ (1 + a/x)^x | e and e^a |
7. Exam Points
- List the seven indeterminate forms and give an example of each — 3-5 marks, almost certain.
- Explain how a 0 × ∞ form is converted to 0/0 or ∞/∞ — 3 marks, frequent.
- Evaluate limits involving 1^∞ and 0⁰ forms using logarithms — 5-7 marks, very frequent.
- State any five standard limits (sin x/x, (1 − cos x)/x², (e^x − 1)/x, (a^x − 1)/x, (1 + x)^(1/x)) — 5 marks, frequent.
- Evaluate a limit by rationalisation or factorisation — 5 marks.
- Explain why "0/0" is called indeterminate, with two examples having different limits — 3 marks.