Limit Calculator

What Is a Limit?

A limit describes the value a function approaches as its input approaches some number. In lim(x→2) (x² − 4)/(x − 2) = 4, the function is undefined at x = 2, yet its values get as close to 4 as you like when x is close to 2: f(1.99) = 3.99 and f(2.01) = 4.01.

That is the key idea: a limit is about the behaviour near a, not the value at a. The function can be undefined at a, or defined with a different value, and the limit can still exist.

How to Find a Limit

Start by substituting x = a. If the result is a number, and the function is built from continuous pieces, that number is the limit. If you get a nonzero number divided by 0, the function has a vertical asymptote and the limit is +∞, −∞, or does not exist, depending on the signs on each side.

If you get 0/0, ∞/∞, ∞ − ∞, 0 · ∞, 1^∞, 0^0 or ∞^0, the form is indeterminate: it gives no answer on its own, and the expression has to be rewritten before the limit can be found.

How the Limit Calculator Works

The calculator first pushes the limit through the expression with the limit laws, substituting wherever the function is continuous. When that produces an indeterminate form, it tries textbook methods in order: factoring and cancelling for rational functions, a common denominator, rewriting |x| on each side, the standard trigonometric limits, the conjugate for square roots, dominant terms and degree comparison at infinity, logarithms for powers, and L'Hôpital's Rule.

The steps shown are the steps it actually used. Each exact answer is also compared with f(x) evaluated close to the target. If the two disagree, or no method applies, the answer is labelled as a numerical estimate rather than presented as exact.

Direct Substitution

Polynomials, rational functions where the denominator is not 0, roots, exponentials, logarithms and trigonometric functions are continuous wherever they are defined. For these, the limit is just the function value: lim(x→3) (x² + 2x + 1) = 9 + 6 + 1 = 16.

Substitution is always the first thing to try. It only fails when it produces division by zero, an indeterminate form, or a value outside the domain.

When Substitution Gives 0/0

The form 0/0 means the numerator and the denominator both approach 0, and their ratio could approach anything. For example, x/x → 1, x²/x → 0 and x/x² has no finite limit at 0, even though all three give 0/0 at x = 0.

So 0/0 is not an answer and not a reason to say the limit does not exist. It is a signal to simplify. Usually it means the numerator and denominator share a factor, often (x − a), that can be cancelled.

How to Solve 0/0 Limits

Factor and cancel: (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2, so the limit as x → 2 is 4.

Multiply by the conjugate when there is a square root: (√(x + 1) − 1)/x becomes x/(x(√(x + 1) + 1)) = 1/(√(x + 1) + 1), so the limit as x → 0 is 1/2.

Use a standard limit for trigonometric expressions, such as sin(x)/x → 1. If none of these apply, L'Hôpital's Rule replaces f/g by f′/g′.

One-Sided Limits

The left-hand limit lim(x→a⁻) f(x) uses only x-values less than a, and the right-hand limit lim(x→a⁺) f(x) uses only values greater than a. The two-sided limit exists exactly when both one-sided limits exist and are equal.

For f(x) = |x|/x, the left-hand limit at 0 is −1 and the right-hand limit is 1. The graph jumps, so lim(x→0) |x|/x does not exist, although each one-sided limit does. One-sided limits are also needed at the edge of a domain: lim(x→0⁺) √x = 0, while √x is not defined to the left of 0.

Limits at Infinity

A limit at infinity describes what f(x) does as x grows without bound. For rational functions, compare the degrees of the numerator and denominator. If the denominator's degree is higher, the limit is 0. If the degrees are equal, the limit is the ratio of the leading coefficients. If the numerator's degree is higher, the function grows without bound.

To see why, divide every term by the highest power of x in the denominator: (3x + 1)/x = 3 + 1/x, and since 1/x → 0, the limit as x → ∞ is 3. A finite limit at infinity is a horizontal asymptote of the graph.

Infinite Limits

When substitution gives a nonzero number divided by 0, the function grows without bound near a, and the graph has a vertical asymptote at x = a. Check the sign on each side: 1/x² is positive on both sides of 0, so lim(x→0) 1/x² = ∞, but 1/x is negative on the left and positive on the right, so its two one-sided limits are −∞ and ∞ and the two-sided limit does not exist.

Writing lim f(x) = ∞ is a precise description of how the limit fails to exist as a real number: the values increase beyond every bound.

Limits That Do Not Exist

A limit does not exist when the left-hand and right-hand limits are different (a jump), when the function grows without bound in different directions on the two sides, when it oscillates without settling, as sin(1/x) does near 0, or when the function is not defined on one side of the point for a two-sided limit.

A limit that does not exist is still a definite answer. It is different from a function value that is undefined: f(a) can be undefined while the limit exists, and the limit can fail to exist even when f(a) is defined.

Important Limit Laws

If lim f(x) = L and lim g(x) = M, then lim [f(x) ± g(x)] = L ± M, lim [c · f(x)] = c · L, lim [f(x) · g(x)] = L · M, and lim [f(x)/g(x)] = L/M provided M ≠ 0. Powers and roots pass through as well: lim [f(x)]ⁿ = Lⁿ.

The Squeeze Theorem covers products with a bounded, oscillating factor: if g(x) ≤ f(x) ≤ h(x) and both g and h approach L, so does f. That is why lim(x→0) x · sin(1/x) = 0.

L'Hôpital's Rule: if f/g gives 0/0 or ∞/∞ and lim f′(x)/g′(x) exists (or is ±∞), then lim f(x)/g(x) equals it. It applies only to those two forms, and other forms must be rewritten as a quotient first.

Common Trigonometric Limits

lim(x→0) sin(x)/x = 1 and lim(x→0) tan(x)/x = 1, with x in radians. More generally, lim(x→0) sin(kx)/(kx) = 1, so lim(x→0) sin(3x)/(5x) = (3/5) · 1 = 3/5.

lim(x→0) (1 − cos x)/x² = 1/2 and lim(x→0) (1 − cos x)/x = 0. Two related exponential and logarithmic limits are lim(x→0) (eˣ − 1)/x = 1 and lim(x→0) ln(1 + x)/x = 1, and the number e itself is a limit: lim(x→∞) (1 + 1/x)ˣ = e.

Limit Examples

lim(x→2) (x² − 4)/(x − 2) = 4, by factoring and cancelling (x − 2).

lim(x→0) (√(x + 1) − 1)/x = 1/2, by multiplying by the conjugate √(x + 1) + 1.

lim(x→∞) (3x² + 2x − 1)/(x² − 5) = 3, since the degrees are equal and the leading coefficients are 3 and 1.

lim(x→0) (x − sin x)/x³ = 1/6, by L'Hôpital's Rule followed by the standard limit (1 − cos x)/x² → 1/2.

lim(x→0) |x|/x does not exist, because the left-hand limit is −1 and the right-hand limit is 1.