Indeterminate Forms: The Limit Formulas to Solve Them

What makes a form indeterminate

A limit like x² as x approaches 3 is no trouble at all: substitution settles it. Things get harder when the calculation lands on an expression like 0/0 or infinity minus infinity, because the symbol’s shape alone says nothing about the result. Treccani reaches this conclusion directly: indeterminate forms — among them infinity minus infinity, zero times infinity, infinity over infinity, and zero over zero — carry that name because it takes comparing which term of the expression dominates the other; reading the written form isn’t enough. Wolfram MathWorld arrives at the same point from another angle: a limit’s value can’t be read from the individual pieces of the expression taken one at a time, and the mathematical community recognizes seven standard indeterminate forms — one to the infinity, 0/0, infinity over infinity, zero times infinity, infinity minus infinity, zero to the zero, and infinity to the zero.

The clearest proof that the written form doesn’t decide the result also comes from MathWorld: two limits can both show up as 0/0 and still have different values, one equal to 1 and one equal to 0.

The concept of a mathematical limit covers what it means for a function to approach a value without ever reaching it; indeterminate forms enter the picture when that approach produces one of the seven forms above and an extra method is needed to finish the calculation.

The limit formulas

Wikipedia describes these limit formulas as a set of results already proved: shortcuts ready for use when direct calculation isn’t straightforward. The table below gathers the ones most used in calculus courses.

Limit	Form	Value
Sine	sin(x)/x, as x approaches 0	1
Cosine	(1 − cos x)/x², as x approaches 0	1/2
Bernoulli (Euler’s number)	(1 + 1/x)^x, as x approaches infinity	e ≈ 2.718
Exponential	(eˣ − 1)/x, as x approaches 0	1
Natural logarithm	ln(1 + x)/x, as x approaches 0	1
Power	((1 + x)ᵃ − 1)/x, as x approaches 0	a

According to Maturando.it, the sine limit is the best known and the first one students learn; its proof doesn’t run through direct calculation but through the squeeze theorem — also called the sandwich theorem — which pins the ratio between two other, easier-to-measure quantities. The English Wikipedia describes the proof as a geometric double inequality: comparing the areas of a smaller triangle, a circular sector, and a larger triangle, all built on the same angle inside a circle of radius 1, shows that the ratio stays between cos(x) and 1, and as x approaches zero both bounds converge to 1.

According to Maturando.it, the exponential and natural logarithm limits aren’t independent of the Bernoulli limit: they’re derived from it. The resulting number e also turns up in exponential equations, where the same constant governs how a quantity grows or decays over time. Theoremz.com reports a generalized version too, useful when the base isn’t e: the limit of (aˣ − 1)/x as x approaches 0 equals the natural logarithm of a, and it reduces to the case above when a = e, because the natural logarithm of e equals 1.

According to Wikipedia, beyond solving a limit outright, these formulas can also replace part of a longer expression with an equivalent infinitesimal, simplifying the calculation before the full formula is even applied.

Two problems, two different paths

Factoring alone already settles some 0/0 forms. Maturando.it gives a classic example: the limit as x approaches 2 of (x² − 4)/(x² − 2x). Both terms vanish at x = 2, but factoring numerator and denominator — (x − 2)(x + 2) on top, x(x − 2) on the bottom — cancels the shared factor (x − 2), and substituting x = 2 into the simplified expression leaves (2 + 2)/2, which is 2.

Worked example: the limit as x approaches 2 of (x² − 4)/(x² − 2x) = the limit as x approaches 2 of (x + 2)/x = 4/2 = 2. No limit formula needed, just factoring and substitution.

When the expression involves trigonometric functions instead, one of these formulas often saves a longer calculation. According to Theoremz.com, the limit as x approaches 0 of sin(x)cos(x)/(2x) is solved by splitting the expression into two factors: sin(x)/x, which is exactly the sine limit and equals 1, and cos(x)/2, which equals 1/2 directly. Multiplying the two results gives the answer.

Worked example: the limit as x approaches 0 of [sin(x)cos(x)]/(2x) = [limit of sin(x)/x] × [limit of cos(x)/2] = 1 × 1/2 = 1/2.