In calculus, finding a limit is usually straightforward: substitute the value and read off the answer. But sometimes substitution produces an expression with no clear value — one that could equal 0, 1, infinity, or anything else depending on the specific functions involved. These are called indeterminate forms.
An indeterminate form is not the same as undefined. It is a signal that direct substitution has failed and that further analysis — algebraic manipulation, L'Hôpital's Rule, or a logarithmic transformation — is needed to find the true limit.
❌ Undefined (not indeterminate)
5 / 0 No limit exists. The expression has no mathematical meaning — it is not a case that can be resolved with calculus techniques.
✅ Indeterminate (resolvable)
lim(x→2) (x²−4)/(x−2) Produces 0/0 under substitution, but the limit does exist and equals 4 once the common factor is cancelled.
There are seven indeterminate forms in calculus. Each arises from a conflict between two competing tendencies, and each has a standard strategy for resolution.
The 7 Indeterminate Forms at a glance
Here is a quick reference guide to the seven forms.
| Type | Symbol | Description |
|---|---|---|
| Quotient | ∞ / ∞ | Numerator and denominator both grow without bound. |
| Quotient | 0 / 0 | Numerator and denominator both shrink to zero. |
| Difference | ∞ - ∞ | Subtraction of two infinite quantities. |
| Product | 0 × ∞ | Multiplication of a tiny number by a huge number. |
| Exponential | 0^0 | Zero raised to a power approaching zero. |
| Exponential | ∞^0 | Infinity raised to a power approaching zero. |
| Exponential | 1^∞ | A value near 1 raised to an infinite power. |
Form by form: what makes each one indeterminate
0 / 0 — the most common form
We know that 0 ÷ 5 = 0, and that 5 ÷ 0 is undefined. But 0 ÷ 0? Consider what division means: a / b = c means b × c = a. If a = 0 and b = 0, then we need 0 × c = 0 — which is true for every value of c. There is no unique answer, so the form is indeterminate. In practice, this is the most frequently encountered form when evaluating limits of rational functions near a point where both numerator and denominator vanish.
Worked Example
Find:
Step 1 — Substitute directly:
Step 2 — Factorise the numerator:
Step 3 — Cancel the common factor:
Step 4 — Substitute:
See also: our full guide to the 0/0 form, including L'Hôpital's Rule and seven worked examples: Zero over zero: Indeterminate Forms
Infinity over Infinity: ∞ / ∞
If both numerator and denominator grow without bound, their ratio could settle at any finite number, or itself diverge to infinity. Which outcome occurs depends entirely on the relative speed of growth of the two functions. For example, x² grows faster than x, so lim(x→∞) x / x² = 0, while lim(x→∞) x² / x = ∞ — same superficial form, opposite results.
The standard resolution is L'Hôpital's Rule: differentiate numerator and denominator separately and re-evaluate. This can be applied repeatedly until the form is resolved.
Detailed treatment with three step-by-step examples using rational functions. Infinity over infinity — worked examples with L'Hôpital's Rule.
0 times Infinity : 0 × ∞ — a tug of war
When one factor approaches zero and the other grows to infinity, there is a direct conflict: the zero factor tries to pull the product down to zero, while the infinite factor tries to push it up without bound. The result depends on which tendency wins.
The technique is to convert the product into a quotient — write f(x) · g(x) as f(x) / (1/g(x)) to get a 0/0 form, or as g(x) / (1/f(x)) to get ∞/∞ — then apply L'Hôpital's Rule.
Worked example — 0 × ∞ form
Find:
As x → 0⁺, we have x → 0 and ln(x) → −∞, giving the form 0 × (−∞).
Step 1 — Rewrite as a quotient to obtain the ∞/∞ form:
Step 2 — Apply L'Hôpital's Rule (differentiate numerator and denominator separately):
Step 3 — Evaluate: 
See also: our full guide to the 0 × ∞ form: Zero times infinity — worked examples
∞ − ∞ — Infinity minus Infinity: subtracting the infinite
It is tempting to assume that ∞ − ∞ = 0, by analogy with 5 − 5 = 0. But infinity is not a fixed number — it is a description of unbounded growth. Two quantities can both grow to infinity at very different rates, so their difference can converge to any constant, diverge, or oscillate.
The standard approach is algebraic manipulation: multiply by a conjugate to convert the subtraction into a quotient, or factor out a common term, then apply L'Hôpital's Rule if needed.
Get more detailed guidance on how to solve ∞ − ∞ forms in calculus.
Exponential indeterminate forms: 0⁰, ∞⁰, and 1^∞
The three exponential forms share the same resolution strategy: take the natural logarithm of the expression, convert the result to a 0/0 or ∞/∞ quotient form, apply L'Hôpital's Rule, then exponentiate the answer to recover the original limit.
0⁰ — the zero-base conflict
We know two things that pull in opposite directions: any non-zero number raised to the power zero equals 1 (so the exponent pushes toward 1), and zero raised to any positive power equals 0 (so the base pushes toward 0). When both base and exponent approach zero simultaneously, neither tendency wins automatically.
- Take the natural log: if L = lim f(x)^g(x), then ln L = lim g(x) · ln f(x)
- This gives a 0 × (−∞) form — convert to a quotient and apply L'Hôpital's Rule
- Exponentiate the result to recover L = e^(answer)
∞⁰ — the zero-exponent conflict
Here, the base grows without bound (which would normally produce ∞), while the exponent shrinks toward zero (which would normally force any base to equal 1). These two tendencies conflict, and the outcome depends on the rates involved.
The resolution is the same as for 0⁰: take the natural log to reduce the exponent to a multiplier, convert to a quotient form, and apply L'Hôpital's Rule.
1^∞ — the most counterintuitive form
This is the form that surprises students most. We intuitively feel that 1 raised to any power is 1. In calculus, though, the base is never exactly 1 — it is approaching 1. A base of 1.000001 raised to a million successive powers does not stay near 1; it can grow to any positive number, famously including e (Euler's number ≈ 2.718), as in:
The resolution again uses the natural log technique: let L = lim f(x)^g(x), compute lim g(x) · ln f(x), and exponentiate the result.
1^∞ Covers the natural log technique in full, including the derivation of e and additional practice examples. Find out more about one to the power of infinity.
Choosing the right resolution strategy
Each indeterminate form has a preferred technique. Choosing the right one upfront avoids unnecessary work.
| Form | First technique to try | If that fails |
|---|---|---|
| 0/0 | Factorise and cancel | Rationalise (conjugate) or L'Hôpital's Rule |
| ∞/∞ | L'Hôpital's Rule | Divide numerator and denominator by the highest-power term |
| 0 × ∞ | Rewrite as 0/0 or ∞/∞ quotient | L'Hôpital's Rule on the rewritten form |
| ∞ − ∞ | Multiply by conjugate or find common denominator | Factor out dominant term, then L'Hôpital |
| 0⁰, ∞⁰, 1^∞ | Take ln L = lim g(x)·ln f(x) | Resolve resulting 0×(−∞) form via L'Hôpital |
Common Mistakes to Avoid
- Confusing indeterminate with undefined. The expression 5/0 is undefined — its limit is ±∞, not a resolvable form. An indeterminate form is an ambiguous form, where the limit can still be found with further work.
- Applying L'Hôpital's Rule without first confirming the form is 0/0 or ∞/∞. The rule only applies to those two quotient forms directly.
- Forgetting to exponentiate after taking the natural log. If you compute ln L, remember that L = e^(ln L) — the final answer must be exponentiated.
- Stopping after one application of L'Hôpital. If the result is still indeterminate, apply the rule again. It can be used repeatedly until the form resolves.
- Assuming ∞ − ∞ = 0 or ∞/∞ = 1. These are the definitions of indeterminate: no algebraic shortcut applies.
For a broader overview of infinity's behaviour in arithmetic and limits, view our guide on properties of infinity in mathematics.
Example Practice Problems
Evaluate 
If we directly apply the limit on the above function, then we will get an indeterminate form of 
because the numerator 
and the denominator 
both are equal to infinity.
Here, we will apply l'Hôpital's Rule because it says that the ratio of the functions is equal to the ratio of their derivatives if their limit approaches to infinity. Hence, to apply l'Hôpital's Rule here, we will take the first derivative of the numerator and the denominator separately.
The first derivative of 
is 
and the derivative of 
is 
. Now, write these derivatives in fractional form like this:
If we apply the limit on the above function, we will still get an indeterminate form of . Hence, now, we will take the second derivative of the above function because it is further differentiable:
The second derivative of the numerator is 
and of the denominator is 
. Write these values in fractional form as shown below:
Simplifying it further will give us the following value:
Evaluate 
If we directly apply the limit on the above function, then we will get an indeterminate form of because the numerator 
\lim_ {x \rightarrow \infty} 21x [/latex] and the denominator 
both are equal to infinity.
Here, we will apply l'Hôpital's Rule because it says that the ratio of the functions is equal to the ratio of their derivatives if their limit approaches to infinity. Hence, to apply l'Hôpital's Rule here, we will take the first derivative of the numerator and the denominator separately. The first derivative of 
is 21 and the derivative of 
is 10. Now, write these derivatives in fractional form like this:
Evaluate 
If we directly apply the limit on the above function, then we will get an indeterminate form of because the numerator 
and the denominator 
both are equal to infinity.
Here, we will apply l'Hôpital's Rule because it says that the ratio of the functions is equal to the ratio of their derivatives if their limit approaches to infinity. Hence, to apply l'Hôpital's Rule here, we will take the first derivative of the numerator and the denominator separately. The first derivative of 
is 
and the derivative of 
is 
. Now, write these derivatives in fractional form like this:
If we apply the limit on the above function, we will still get an indeterminate form of . Hence, we will take the second derivative of the above function because it is further differentiable:
The second derivative of the numerator is 
and of the denominator is 
. Write these values in fractional form as shown below:
Simplifying it further will give us the following value:
Evaluate 
.
If we directly apply the limit on the above function, then we will get an indeterminate form of because the numerator 
and the denominator 
both are equal to infinity.
Here, we will apply l'Hôpital's Rule because it says that the ratio of the functions is equal to the ratio of their derivatives if their limit approaches to infinity. Hence, to apply l'Hôpital's Rule here, we will take the first derivative of the numerator and denominator separately. The first derivative of 
is 
and the derivative of 
is 
. Now, write these derivatives in fractional form like this:
Evaluate 
.
If we directly apply the limit on the above function, then we will get an indeterminate form of because the numerator and the denominator 
both are equal to infinity.
Here, we will apply l'Hôpital's Rule because it says that the ratio of the functions is equal to the ratio of their derivatives if their limit approaches to infinity. Hence, to apply l'Hôpital's Rule here, we will take the first derivative of the numerator and the denominator separately. The first derivative of is and the derivative of 
is 
. Now, write these derivatives in fractional form like this:
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what will be sign of -infinity is to power infinity plus or minus
The sign **cannot be determined**.
For a negative base:
* **Even exponent** → **positive**
* **Odd exponent** → **negative**
So as the exponent tends to infinity, the sign depends on whether the exponent is even or odd. Therefore, ((-\infty)^\infty) has **no definite sign**—it can be **positive or negative**.