Calculator

Limit Calculator with Steps

Evaluate one-sided limits, limits at infinity, and every indeterminate form, with the factoring, conjugate, or L'Hôpital step written out rather than assumed.

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How to use it

State the expression and the value the variable approaches, including the side if you need one: “as x approaches 3 from the left”. For a limit at infinity, say infinity rather than a large number. If your course has not covered L'Hôpital's rule yet, say so in the question and the solution stays algebraic. Limits are the foundation the calculus solver builds derivatives on, so the same input works there for a longer problem.

The method behind it

A limit describes where a function is heading as its input approaches some value, and it deliberately ignores what happens at that value. That is the whole reason limits exist: $(x^{2} - 4) / (x - 2)$ has no value at $x = 2$ , but it is heading somewhere perfectly definite.

The two-sided limit exists only when both one-sided limits exist and agree. At a jump discontinuity they do not, and no amount of algebra will manufacture one.

x \to a lim f (x) = L ⟺ x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = L

Always try substitution first. If the function is continuous at $a$ , the limit is just $f (a)$ and there is nothing to do. Substitution earns its keep even when it fails, because what it produces tells you which technique comes next.

If substitution returns one of these, the expression has not failed — it has simply not answered yet.

\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, 0 \cdot \infty, 1^{\infty}

For $0/0$ in a rational function, factor and cancel: the cancelled factor is exactly the hole. For a difference of square roots, multiply by the conjugate, which converts the subtraction into a difference of squares and clears the radical from where it was causing trouble. For a compound fraction, combine over a common denominator first. Being able to factor a quadratic on sight is what makes the first of these fast.

L'Hôpital's rule applies only to $0/0$ and $\infty/\infty$ . Given those, and with both functions differentiable near $a$ , you may differentiate the numerator and the denominator separately — not as a quotient — and try again.

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}

For limits at infinity, divide every term by the highest power appearing in the denominator. Each term of lower degree then visibly collapses to zero, and what is left gives the answer directly: the ratio of the leading coefficients when the two degrees match, zero when the numerator has the lower degree, and no finite limit when it has the higher one. Two standard limits are worth knowing cold because they resolve most trigonometric indeterminates without any rule at all.

x \to 0 lim \frac{sin x}{x} = 1, x \to 0 lim \frac{1 - cos x}{x} = 0

The wider picture — what continuity requires, and how limits connect to derivatives — is set out in limits and continuity, intuition first.

Worked examples

Worked example

lim_{x \to 4} \frac{x - 2}{x - 4}

1
Substitute first to identify the form. Both parts vanish, so this is 0/0 and the expression is still in play.
$\frac{4 - 2}{4 - 4} = \frac{0}{0}$
2
Multiply top and bottom by the conjugate of the numerator. This is multiplication by 1, so the function is unchanged.
$\frac{x - 2}{x - 4} \cdot \frac{x + 2}{x + 2}$
3
The numerator becomes a difference of squares and the radical disappears from where it was blocking the cancellation.
$\frac{x - 4}{( x - 4 ) ( x + 2 )}$
4
Cancel the common factor. This is legal because x is approaching 4, never equal to it, so the factor is never zero.
$\frac{1}{x + 2}$
5
Now substitution works, because what is left is continuous at 4.
$\frac{1}{4 + 2} = \frac{1}{4}$

Answer

lim_{x \to 4} \frac{x - 2}{x - 4} = \frac{1}{4}

Worked example

lim_{x \to \infty} (x^{2} + 3 x - x)

1
Both terms grow without bound, so this is the indeterminate difference infinity minus infinity. The two growths are nearly equal, and the answer is what remains of the gap.
$\infty - \infty$
2
There is no fraction yet, so make one: multiply and divide by the conjugate.
$(x^{2} + 3 x - x) \cdot \frac{x ^{2} + 3 x + x}{x ^{2} + 3 x + x}$
3
The numerator collapses, since the squares cancel exactly.
$\frac{( x ^{2} + 3 x ) - x ^{2}}{x ^{2} + 3 x + x} = \frac{3 x}{x ^{2} + 3 x + x}$
4
Divide top and bottom by x. Inside the radical that means dividing by x squared, because x is positive as it heads to infinity.
$\frac{3}{1 + \frac{3}{x} + 1}$
5
The 3/x term goes to zero, leaving a value that is finite even though both original terms were not.
$\frac{3}{1 + 1} = \frac{3}{2}$

Answer

lim_{x \to \infty} (x^{2} + 3 x - x) = \frac{3}{2}

One limit, two routes

Most indeterminate forms can be resolved either algebraically or with L'Hôpital's rule, and seeing the same limit both ways tells you which one your particular problem rewards. Work through this one a step at a time before revealing each line.

Evaluate the limit of (1 − cos x)/x² as x → 0, twice

0 of 7 steps

Try it yourself first — you'll remember it longer.

Common mistakes

Using L'Hôpital on a form that is not indeterminate

The rule requires $0/0$ or $\infty/\infty$ . Applying it to $lim_{x \to 0} \frac{x + 1}{x + 2}$ gives $1/1 = 1$ , but substitution shows the answer is $1/2$ . Check the form before every application, including repeat applications.

Differentiating as a quotient

L'Hôpital's rule differentiates the numerator and the denominator independently. It is not the quotient rule and produces a different expression entirely. If you see a squared denominator appear, you have applied the wrong rule.

Confusing the limit with the function value

A function can have a limit at a point where it is undefined, and it can be defined at a point where the limit disagrees with the value. Those two things matching is the definition of continuity, not something you may assume while computing.

Practice

Answers are checked here — nothing is sent anywhere.

1
$lim_{x \to 2} \frac{x ^{2} - 4}{x - 2}$
2
$lim_{x \to 0} \frac{s i n 5 x}{x}$
3
$lim_{x \to \infty} \frac{4 x ^{3} + 2 x}{7 x ^{3} - x ^{2}}$

Frequently asked questions

Can it do one-sided limits?

Yes. Write the limit as x approaches 2 from the left, or use the notation x → 2⁻. One-sided limits are the only way to describe behaviour at a jump or at the edge of a domain, so they are worth asking for explicitly.

Does it use L'Hôpital's rule or algebra?

Whichever the problem calls for, and it says which. Many courses require the algebraic route because L'Hôpital's rule is introduced later, so the working shows the factoring or conjugate step when one exists.

How are limits at infinity handled?

By dividing through by the highest power in the denominator, which turns every remaining term into something that visibly goes to zero. The answer explains why the leading terms dominate rather than just quoting a rule about degrees.

What if the limit does not exist?

The solution says so and shows why: the one-sided limits disagree, the function oscillates without settling, or it grows without bound. Writing the answer as infinity is a description of the failure, not a value.

What does it cost?

The first solution is free and requires no account. Free accounts get three a day. Gauth Plus is $11.99 a month with a three-day free trial and removes the daily limit.

Keep going

Calculus solverLimits, derivatives, integrals, and series with the reasoning at each line.Limits and continuity, intuition firstWhat a limit means, and the three conditions a function must meet to be continuous.Factoring quadratics: the four patternsThe skill that makes factor-and-cancel limits fast instead of fiddly.Derivative calculatorThe limit of a difference quotient, evaluated for every standard function.