The Seven Indeterminate Forms

A limit that gives 0/0 tells you nothing. The limit might be 5, 0, ∞, or it might not exist. That is what indeterminate means: the form alone does not determine the answer. Seven indeterminate forms exist: 0/0, ∞/∞, 0·∞, ∞−∞, 0^0, 1^∞, and ∞^0. Every other limit form, like 0/∞, ∞/0, or ∞+∞, is determinate and gives an immediate answer. Recognise the form and pick the fix.

A limit is the value a function approaches as the input gets arbitrarily close to a point. When you plug the target x-value in and get 0/0, the result is a hole in the algebra, not the final answer. The same is true for ∞/∞, 0·∞, ∞−∞, 0^0, 1^∞, and ∞^0. These seven patterns are the complete indeterminate form list (Stewart §4.4, §3.3). Any other combination of zero, infinity, or a finite number gives a determinate result.

For example, 0/∞ equals 0. ∞/0 equals ∞ or −∞. ∞+∞ equals ∞. ∞·∞ equals ∞. A finite number divided by ∞ equals 0. These are not indeterminate; they are settled by the rules for infinite limits and the limit laws for finite limits (Stewart §2.3). The only trouble comes from the seven forms listed above.

The two most common indeterminate forms are 0/0 and ∞/∞. Both respond to the same fix: L'Hôpital's Rule. L'Hôpital's Rule applies only when the limit is exactly 0/0 or ∞/∞, and the limit of the quotient of derivatives exists or is ±∞ (Stewart §4.4). Differentiate the numerator and denominator separately, then take the limit again.

0/0 Case: lim_{x→0} sin x / x

Direct substitution gives 0/0. Apply L'Hôpital's Rule: derivative of sin x is cos x, derivative of x is 1. The limit becomes lim_{x→0} cos x / 1 = 1. This limit is the classic squeeze theorem case as well (Stewart §2.3, §3.3), but L'Hôpital works in one step. Check the conditions: the original limit must be indeterminate and the derivative limit must exist.

∞/∞ Case: lim_{x→∞} ln x / x

Direct substitution gives ∞/∞. Apply L'Hôpital: derivative of ln x is 1/x, derivative of x is 1. The limit becomes lim_{x→∞} (1/x) / 1 = 0. The logarithm grows slower than any positive power of x, so the ratio goes to zero. If L'Hôpital gives another ∞/∞ or 0/0, apply it again. Multiple applications are valid as long as the form remains indeterminate.

The 0·∞ and ∞−∞ forms are not directly solvable by L'Hôpital. Rewrite them as a quotient form first.

0·∞ Case: lim_{x→0+} x ln x

Direct substitution gives 0·(−∞). Rewrite x ln x as ln x / (1/x). As x→0+, ln x → −∞ and 1/x → ∞, so the quotient is −∞/∞, an ∞/∞ form. Apply L'Hôpital: derivative of ln x is 1/x, derivative of 1/x is −1/x². The limit becomes lim_{x→0+} (1/x) / (−1/x²) = lim_{x→0+} (−x) = 0. The standard technique for 0·∞ is to rewrite as 0/(1/∞) or ∞/(1/0) to obtain 0/0 or ∞/∞ (Stewart §4.4).

∞−∞ Case: lim_{x→0+} (1/x − 1/sin x)

Direct substitution gives ∞−∞. Combine the fractions over a common denominator: (sin x − x) / (x sin x). As x→0+, numerator and denominator both approach 0, giving 0/0. Apply L'Hôpital: derivative of sin x − x is cos x − 1, derivative of x sin x is sin x + x cos x. At x=0, that gives 0/0 again. Apply L'Hôpital a second time: derivative of cos x − 1 is −sin x, derivative of sin x + x cos x is 2 cos x − x sin x. At x=0, −sin 0 = 0, 2 cos 0 − 0 = 2, so the limit is 0/2 = 0. The alternative fix for ∞−∞ is to rationalize or find a common denominator (Stewart §4.4).

The three exponent forms, 0^0, 1^∞, and ∞^0, are handled by a single method: apply the natural logarithm to bring the exponent down, evaluate the limit of the logarithm, then exponentiate the result (Stewart §4.4). This converts the exponent form into a product form, which can then be rewritten as a quotient.

1^∞ Case: lim_{x→0} (1+x)^(1/x)

Direct substitution gives 1^∞. Let y = (1+x)^(1/x). Take ln y = (1/x) ln(1+x). As x→0, ln(1+x) → 0, so the product is 0·∞. Rewrite as ln(1+x)/x, which is 0/0. Apply L'Hôpital: derivative of ln(1+x) is 1/(1+x), derivative of x is 1. The limit of ln y is lim_{x→0} 1/(1+x) = 1. Then y = e^1 = e. This limit defines the number e.

0^0 Case: lim_{x→0+} x^x

Direct substitution gives 0^0. Let y = x^x. Take ln y = x ln x. As x→0+, this is the 0·∞ form from earlier, which we solved as 0. So ln y → 0, and y = e^0 = 1.

∞^0 Case: lim_{x→∞} x^(1/x)

Direct substitution gives ∞^0. Let y = x^(1/x). Take ln y = (1/x) ln x. As x→∞, ln x → ∞ and 1/x → 0, so the product is ∞·0. Rewrite as ln x / x, which is ∞/∞. Apply L'Hôpital: derivative of ln x is 1/x, derivative of x is 1. The limit is 0. Then y = e^0 = 1.

Not every zero-or-infinity combination is indeterminate. The following forms are determinate and require no special technique beyond direct substitution or the rules for infinite limits.

1/0

A finite number divided by zero yields an infinite limit: ∞ or −∞, depending on sign and direction. For example, lim_{x→0} 1/x² = ∞ (Stewart §2.2). This is not an indeterminate form; it is a vertical asymptote. The limit does not exist as a finite number, but the notation lim = ∞ is accepted in AP Calculus and Stewart to indicate unbounded growth.

∞+∞

∞ plus ∞ is ∞. No conflict. For example, lim_{x→∞} (x² + x) = ∞. Similarly, ∞·∞ = ∞, and a finite number times ∞ is ∞ (with sign).

0/∞ and ∞/0

0/∞ = 0. ∞/0 = ∞ or −∞. Both are settled immediately. A common mistake is to treat 0/∞ as indeterminate because it contains zero and infinity, but the zero in the numerator dominates.

If you ever get a form that is not one of the seven, do not reach for L'Hôpital. L'Hôpital only applies to 0/0 or ∞/∞, and applying it to a determinate form can give a wrong answer. The failure case is applying L'Hôpital to 0/∞, which is 0, or to ∞/0, which is infinite.

The table below summarises the seven indeterminate forms, a canonical example for each, and the technique to resolve it.

Form	Example	Technique
0/0	lim_{x→0} sin x / x	L'Hôpital's Rule (or factoring/rationalizing)
∞/∞	lim_{x→∞} ln x / x	L'Hôpital's Rule
0·∞	lim_{x→0+} x ln x	Rewrite as 0/(1/∞) or ∞/(1/0) to get 0/0 or ∞/∞
∞−∞	lim_{x→0+} (1/x − 1/sin x)	Combine fractions or rationalize to get 0/0 or ∞/∞
0^0	lim_{x→0+} x^x	Apply ln, evaluate product limit, then exponentiate
1^∞	lim_{x→0} (1+x)^(1/x)	Apply ln, evaluate product limit, then exponentiate
∞^0	lim_{x→∞} x^(1/x)	Apply ln, evaluate product limit, then exponentiate

The most frequent error is applying L'Hôpital's Rule to a limit that is not 0/0 or ∞/∞.But lim_{x→0} (1) / (1/x) is 1/∞, which equals 0, and L'Hôpital would produce 0/0, a false result. Check the form first.

Another failure mode is forgetting to check one-sided limits. A limit exists only if the left and right limits are equal and finite. For infinite limits, the direction matters: lim_{x→0} 1/x does not exist because the left limit is −∞ and the right limit is +∞. The same applies to piecewise functions and absolute values.

Confusing a 0/0 limit with a removable discontinuity is common. They are the same thing: a 0/0 form indicates a hole that can be filled by simplifying. The limit exists and is finite, even if the function is not defined at the point. The difference between a removable discontinuity and a jump discontinuity is that in a jump, the left and right limits are different finite numbers, so the two-sided limit does not exist.

L'Hôpital's Rule is the most powerful tool for 0/0 and ∞/∞, but not every problem requires it. Factoring works when a common factor causes the zero in numerator and denominator. Rationalizing works when a radical is present. The squeeze theorem is the right tool when the function oscillates or is bounded between two others with the same limit.

For example, lim_{x→2} (x²−4)/(x−2) is 0/0. Factor (x²−4) into (x−2)(x+2), cancel (x−2), and the limit is 4. L'Hôpital also works: derivative of numerator is 2x, derivative of denominator is 1, limit is 4. Both are valid, but factoring is faster and avoids the derivative condition.

The squeeze theorem is necessary for limits like lim_{x→0} x² sin(1/x). Direct substitution gives 0·sin(∞), which is 0·(oscillating). The squeeze theorem shows the limit is 0 because −x² ≤ x² sin(1/x) ≤ x², and both bounds go to 0.

The single most common mistake is misidentifying the form. Students see a zero in the denominator and immediately assume the limit is infinite, missing a 0/0 that could be cancelled. Or they see ∞/∞ and apply L'Hôpital without checking that the limit of the derivative quotient exists. Or they treat 1^∞ as equal to 1, which is false, as the case (1+x)^(1/x) shows, it equals e.

Always write down the form before choosing a technique. If the form is one of the seven, you have work to do. If it is any other combination, the answer is immediate. Recognise the form and pick the fix.