What makes a limit indeterminate
A quotient f(x)/g(x) whose numerator and denominator both tend to zero is called indeterminate because the value of the limit is not decided by that fact alone. Compare three examples at x → 0: x/x → 1, x²/x → 0, and x/x² → ∞. All three are of the form 0/0 and all three behave differently. What settles the answer is the rate at which each part approaches zero, and rate is exactly what a derivative measures.
That is the whole content of L'Hôpital's rule. If both functions vanish at a, then near a they are approximated by f(a) + f'(a)(x−a) and g(a) + g'(a)(x−a), and with both constant terms zero the quotient reduces to f'(a)/g'(a). When that ratio is itself 0/0, both first-order terms vanish too and you go one order deeper. The rule is a systematic way of comparing Taylor expansions without writing them out.
Two other forms are genuinely indeterminate and covered by the same rule: ∞/∞, and the closely related family 0·∞, ∞−∞, 0⁰, 1^∞ and ∞⁰, which are first rewritten as a quotient or handled by taking logarithms. Everything else — including 1/0, 0/1 and ∞+∞ — is not indeterminate and must not be passed through the rule.
This calculator works at a finite approach point. For a limit at infinity, substitute x = 1/t and take t → 0⁺: for example lim(x→∞) x/eˣ becomes lim(t→0⁺) (1/t)/e^(1/t), which is easier to reason about but still needs care. The rule itself is valid at infinity; it is only this page's numerical machinery that needs a finite point.
The rule, its hypotheses, and how this page applies it
The precise statement is: if f and g are differentiable on an interval around a (except possibly at a itself), if g'(x) ≠ 0 there, and if f(x) → 0 and g(x) → 0 as x → a, then lim f/g = lim f'/g' provided the second limit exists. Every clause matters. The rule is proved from the Cauchy mean value theorem, which is why it needs differentiability on a whole neighbourhood rather than just at the point.
The last clause is the one people forget. The rule says the limit of the derivative quotient, if it exists, equals the original limit — not the converse. For lim(x→∞) (x + sin x)/x, the derivative quotient (1 + cos x)/1 has no limit, but the original limit is plainly 1. A failure of the rule is not evidence that the original limit fails to exist.
The calculator applies the rule as a chain on derivative values at the point. It first tests whether the form is genuinely 0/0, using a threshold scaled to the size of each function nearby, so a value that is zero only to rounding is treated as zero. If it is indeterminate, it computes f⁽ᵏ⁾(a) and g⁽ᵏ⁾(a) for k = 1, 2, 3, … by central differences with Richardson extrapolation, and stops at the first order where the denominator derivative is non-zero. The limit is then that ratio.
Alongside it, the page computes the limit a second way, with no differentiation at all: it samples the quotient at two or three decreasing offsets from a and extrapolates the sequence. The two methods have completely different failure modes, so agreement between them is real evidence. Disagreement is flagged, and it usually means a high-order derivative has lost precision — the fourth numerical derivative of a function is a fragile object.
Worked example: lim(x→0) (1 − cos x)/x²
Do it by hand and follow the table above.
- Check the form. At x = 0, the numerator is 1 − cos 0 = 1 − 1 = 0 and the denominator is 0² = 0. This is 0/0, so the rule applies.
- Differentiate separately. f'(x) = sin x and g'(x) = 2x. Do not use the quotient rule — the rule differentiates the two functions independently.
- Test again. At x = 0, f'(0) = sin 0 = 0 and g'(0) = 0. Still 0/0, so apply the rule a second time.
- Differentiate again. f''(x) = cos x and g''(x) = 2.
- Evaluate. f''(0) = 1 and g''(0) = 2, and 2 is not zero, so the chain stops. The limit is 1/2 = 0.5.
- Cross-check with the series. cos x = 1 − x²/2 + x⁴/24 − …, so 1 − cos x = x²/2 − x⁴/24 + … and the quotient is 1/2 − x²/24 + …, which tends to 1/2. It also tells you the approach is from below, at a rate proportional to x².
The series cross-check is worth doing every time you can. Sample the quotient directly to see it: at x = 0.1 the value is (1 − 0.995004165)/0.01 = 0.4995835, and at x = 0.05 it is 0.4998958. The gap from 0.5 shrinks from 0.0004165 to 0.0001042 — a factor of four for a halving of x, exactly the x²/24 behaviour the series predicts: 0.01/24 = 0.0004167. Two independent routes to the same number is what confidence in a limit looks like.
Reading the output
The first thing to read is the number of applications. Zero means the quotient was never indeterminate — the denominator did not vanish — and the limit came from substitution. That is a correct answer, and it also means using L'Hôpital's rule on that problem would have produced a different and wrong number. Applying the rule to a determinate form is the most common misuse of it, and it is what the warning on that case is there to prevent.
One application is the ordinary case. Two or three are common for problems built from even functions or from series whose leading terms cancel. Four or more should make you suspicious: it may be genuine, but every extra order costs numerical accuracy, and by the fourth derivative the difference formula is working with function values that agree in most of their digits.
Then compare the limit with the independent estimate. The gap between them is reported; on the worked example it is below 10⁻⁹. A gap comparable with the limit itself means one of the two methods has failed, and the table of orders usually shows which — a derivative pair that both look like small multiples of rounding noise is the tell.
Finally read the chart. It plots the quotient itself on both sides of a, so a limit that exists appears as a curve running smoothly into a single height from both directions. A jump between the two sides means the two one-sided limits differ and the two-sided limit does not exist, in which case you should re-run with a direction selected. A curve that runs off the top means the quotient is unbounded and there is no finite limit at all.
Reference: standard indeterminate limits
| Limit | Form | Applications | Value |
|---|---|---|---|
| sin x / x, x → 0 | 0/0 | 1 | 1 |
| (eˣ − 1) / x, x → 0 | 0/0 | 1 | 1 |
| ln x / (x − 1), x → 1 | 0/0 | 1 | 1 |
| (1 − cos x) / x², x → 0 | 0/0 | 2 | 1/2 |
| (eˣ − 1 − x) / x², x → 0 | 0/0 | 2 | 1/2 |
| (x − sin x) / x³, x → 0 | 0/0 | 3 | 1/6 |
| tan x / x, x → 0 | 0/0 | 1 | 1 |
| (x² − 4) / (x − 2), x → 2 | 0/0 | 1 | 4 |
Every value in this table is also the leading Taylor coefficient ratio. For (x − sin x)/x³, the series x³/6 − x⁵/120 + … over x³ gives 1/6 directly, which is why three applications are needed: the first two orders both vanish.
Mistakes that produce wrong limits
- Using the quotient rule instead of the rule. L'Hôpital differentiates numerator and denominator separately. Differentiating f/g as a single expression is a different calculation and gives a different, wrong answer.
- Applying it to a determinate form. lim(x→0) sin x/(x+1) is 0/1 = 0 by substitution. Differentiating gives cos x/1 → 1, which is wrong. Always verify the form first.
- Continuing after the form resolves. Once the denominator derivative is non-zero, stop. Another application changes the answer.
- Forgetting that the rule can fail without the limit failing. If f'/g' has no limit, you learn nothing about f/g; find another method.
- Ignoring the direction. ln x exists only for x > 0, so any limit involving it at 0 is one-sided. A two-sided request there has no answer.
- Using the rule where a factorisation is faster. (x² − 4)/(x − 2) is x + 2 for x ≠ 2, so the limit is 4 in one line without any calculus.
- Trusting a fourth-order numerical derivative. Each extra order costs several significant figures. Cross-check against the series or against the independent estimate.
Rewriting the other indeterminate forms
0·∞: write f·g as f/(1/g) to make it 0/0, or g/(1/f) to make it ∞/∞ — pick whichever gives derivatives you can handle. ∞−∞: combine over a common denominator. 1^∞, 0⁰ and ∞⁰: take logarithms, so y = f^g becomes ln y = g·ln f, resolve that as a 0·∞ form, and exponentiate at the end. That last route is how lim(x→0⁺) xˣ = 1 is established. To see how quickly a series representation converges once you have one, the radius of convergence calculator tells you where it is valid at all.
Alternatives to the rule, and where it sits
Taylor series are the more powerful tool for most 0/0 limits, and often the faster one. Expand numerator and denominator to a few terms, cancel the common power, and read off the ratio of the leading coefficients — that is the same answer L'Hôpital produces, obtained in one step instead of k. It also tells you how fast the quotient approaches its limit, which the rule does not. Where a series is valid is a separate question, answered by the radius and interval of convergence calculator.
Algebra beats both when it applies. Factorising, rationalising a surd by multiplying by its conjugate, or dividing through by the highest power are all exact and instant. Reach for the rule when those fail.
The derivatives the rule needs are the same ones the numerical derivative calculator computes, and its discussion of step size and cancellation explains directly why the chain here becomes unreliable at high order: round-off in a k-th difference grows like ε/hᵏ. Where the limit you are chasing is the derivative itself — the difference quotient at a point — the two pages are computing the same object from opposite directions.
Finally, limits of this kind are how continuous extensions are built. A function such as sin(x)/x, undefined at the origin, becomes continuous everywhere once you define its value there to be the limit 1. That extension is what lets you integrate it — the Riemann sum calculator will happily sample the extended function — and what makes the sinc function a well-behaved object in signal processing.