Limit Calculator

Q: What does the limit calculator do when substitution gives 0/0?

It recognises the indeterminate form and applies L'Hôpital's rule, differentiating the numerator and the denominator separately and substituting again, repeating until the form is resolved or the method stops helping.

Q: How do I find a one-sided limit?

Choose 'From the right' for x → a⁺ or 'From the left' for x → a⁻. One-sided limits matter at vertical asymptotes, at endpoints of a domain such as √x at 0, and for piecewise or absolute-value expressions.

Q: How do I enter a limit at infinity?

Type inf, or -inf for negative infinity, in the 'x approaches' box. The calculator uses leading-term comparisons, L'Hôpital's rule and exponential rewriting as needed.

Q: What does 'the limit does not exist' mean here?

Either the left-hand and right-hand limits are different, which the calculator states with both values, or the values oscillate and never settle on one number or on ±∞.

Q: Is a limit of infinity the same as no limit?

Strictly, an infinite limit is not a real number, so the limit does not exist in the real numbers. Saying it is +∞ is more precise, because it describes how the values behave, so the calculator reports it that way.

Q: Why does the calculator check limits numerically?

Symbolic manipulation can go wrong in subtle cases. Evaluating the function on a sequence of points approaching the target confirms the answer, and if the numbers contradict the symbolic result, no answer is shown.

Q: Can it evaluate limits of the form 1 to the power infinity?

Yes. It takes logarithms to turn the power into a product, resolves the resulting 0·∞ form with L'Hôpital's rule and exponentiates, which is how (1 + 1/x)^x → e is shown.

Q: Does the limit equal the function value at that point?

Only when the function is continuous there. For (x² − 4)/(x − 2) the limit at 2 is 4 even though the expression is undefined at x = 2.

Q: Can it handle limits with two variables?

No. The calculator works with one variable, normally x. Multivariable limits depend on the path of approach and need different techniques.

About the Limit Calculator

Limits are the foundation that the rest of calculus is built on. Derivatives are limits of difference quotients, integrals are limits of sums, and the convergence of a series is a limit of partial sums. Yet limit problems are often where students first meet expressions like $\frac00$ that look meaningless, and it is easy to lose track of which technique applies.

This calculator works through a limit the way a careful student would. It first tries direct substitution. If that produces an indeterminate form, it names the form — $\frac00$ , $\frac{\infty}{\infty}$ , $0\cdot\infty$ , $\infty - \infty$ , $1^\infty$ , $0^0$ or $\infty^0$ — and applies the matching method: L'Hôpital's rule, rewriting a product as a quotient, multiplying by a conjugate or taking logarithms. Each step is shown with the derivatives that were taken.

Every symbolic answer is then compared against the actual values of the function at points closer and closer to the target. If the two disagree, no answer is shown, which protects you from the confident wrong results that symbolic tricks can occasionally produce. The calculator also handles one-sided limits and tells you when a two-sided limit does not exist because the two sides disagree.

What is a limit?

The limit of $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ gets arbitrarily close to when $x$ is close to, but not equal to, $a$ . It describes the trend of the function near a point, not its value at that point. The function does not even need to be defined at $a$ : $\frac{\sin x}{x}$ is undefined at 0, yet its limit there is exactly 1.

Formally, $\lim_{x\to a} f(x) = L$ means that for every tolerance $\varepsilon > 0$ there is a distance $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 < |x - a| < \delta$ . In plain words, you can force $f(x)$ as close to $L$ as you like by keeping $x$ close enough to $a$ .

A limit can also be infinite, written $\lim f(x) = \infty$ , when the values grow without bound, and $x$ can approach $\pm\infty$ itself, which describes the long-run behaviour of the function and its horizontal asymptotes. A two-sided limit exists only when the left-hand and right-hand limits exist and are equal.

How to use the Limit Calculator

Type the function of x, for example sin(x)/x, (x^2-4)/(x-2) or (1+1/x)^x.
Enter the value x approaches: a number, a fraction, pi, inf or -inf.
Choose two-sided, from the right (x → a⁺) or from the left (x → a⁻).
Press Find limit to see the substitution attempt, the indeterminate form and each resolving step.
Check the final line, which states the limit, ±∞ or that the limit does not exist and why.

Formulas

L'Hôpital's rule

\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\quad\text{for } \tfrac00 \text{ or } \tfrac{\infty}{\infty}

Fundamental trigonometric limit

\lim_{x\to 0}\frac{\sin x}{x} = 1

Definition of e

\lim_{x\to\infty}\left(1 + \frac1x\right)^x = e

Exponential forms

\lim f^{\,g} = e^{\lim g\,\ln f}

Two-sided limit

\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L

Worked example: sin(x)/x, x → 0

The worked example below is generated live for $\lim_{x\to 0} \frac{\sin x}{x}$ . Substituting $x = 0$ gives $\frac00$ , which is indeterminate, so the calculator applies L'Hôpital's rule: the derivative of the numerator is $\cos x$ and the derivative of the denominator is 1. Substituting again gives $\cos 0 = 1$ , so the limit is 1.

A table of values agrees. At $x = 0.1$ the ratio is about 0.99833 and at $x = 0.01$ it is about 0.9999833, closing in on 1 from below. This particular limit is the reason the derivative of $\sin x$ is $\cos x$ , so it is worth recognising on sight.

The solution below is generated by the same engine that powers the calculator above.

Final answer

\lim_{x \to 0} \frac{\sin\left( x\right)}{x} = 1

How we read your problem

Find the limit of sin(x) / x as x → 0.

\lim_{x \to 0} \frac{\sin\left( x\right)}{x}

Step-by-step solution

Indeterminate form 0/0 — apply L'Hôpital's rule
Differentiate the numerator and the denominator separately (this is not the quotient rule).
$\lim_{x \to 0} \frac{\sin\left( x\right)}{x} = \lim_{x \to 0} \frac{\cos\left( x\right)}{1}$
Simplify
$\lim_{x \to 0} \frac{\cos\left( x\right)}{1} = \lim_{x \to 0} \cos\left( x\right)$
Direct substitution
The expression is continuous at x = 0, so the limit equals its value there.
$\lim_{x \to 0} \cos\left( x\right) = \cos\left(0\right) = 1$

Formulas used

L'Hôpital's rule
$\lim \frac{f}{g} = \lim \frac{f'}{g'}\ \text{for } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty}$
Two-sided limit
$\lim_{x\to a} f = L \iff \lim_{x\to a^-} f = \lim_{x\to a^+} f = L$

A limit describes the value a function approaches, not necessarily the value it takes. When substitution gives an indeterminate form, the expression is rewritten (L'Hôpital's rule, a conjugate, or logarithms) until substitution works.

Verification

Numerical values from the right agree: f(0.01) = 0.99998333, f(0.001) = 0.99999983, f(0.0001) = 1, f(0.00001) = 1 ✓
Numerical values from the left agree: f(-0.01) = 0.99998333, f(-0.001) = 0.99999983, f(-0.0001) = 1, f(-0.00001) = 1 ✓

Detailed explanation

Start with direct substitution

For a function that is continuous at the point, the limit is simply its value there, so the first move is always to substitute. Polynomials, exponentials, sine and cosine are continuous everywhere, and rational functions are continuous wherever the denominator is not zero. If substitution produces an ordinary number, you are done.

Substitution can also reveal an infinite limit. A nonzero number divided by something approaching zero, such as $\frac1x$ near 0, grows without bound. The sign then depends on the side: the calculator reports $+\infty$ from the right and $-\infty$ from the left, and therefore that the two-sided limit does not exist.

Indeterminate forms and L'Hôpital's rule

Forms such as $\frac00$ and $\frac{\infty}{\infty}$ are called indeterminate because the form alone says nothing about the answer. $\frac{x^2}{x}$ , $\frac{x}{x}$ and $\frac{x}{x^2}$ all give $\frac00$ at zero, yet their limits are 0, 1 and nonexistent. You have to look more closely at how fast the numerator and denominator approach their limits.

L'Hôpital's rule does exactly that by comparing rates of change. When both parts tend to 0, or both tend to infinity, the limit of the ratio equals the limit of the ratio of derivatives, provided that new limit exists. For $\frac{x^2-4}{x-2}$ at $x = 2$ the derivatives are $2x$ and 1, so the limit is 4. Factoring and cancelling gives the same answer, $x + 2 \to 4$ , which is a good way to check.

The rule may need to be applied more than once. $\frac{1-\cos x}{x^2}$ is still $\frac00$ after one round; a second round gives $\frac{\cos x}{2}$ and the limit $\frac12$ . The calculator repeats the rule automatically, stopping if the expressions start to grow instead of simplify.

Products, differences and powers

L'Hôpital's rule only applies to quotients, so other indeterminate forms must be rewritten first. A product $0 \cdot \infty$ becomes a quotient by moving one factor into the denominator as its reciprocal. A difference $\infty - \infty$ involving a square root, such as $\sqrt{x^2 + x} - x$ at infinity, is multiplied by its conjugate; that one simplifies to a limit of $\frac12$ .

The exponential forms $1^\infty$ , $0^0$ and $\infty^0$ are handled with logarithms. To evaluate $\lim (1 + \frac1x)^x$ , take the natural logarithm to get $x \ln(1 + \frac1x)$ , a $\infty\cdot 0$ form, rewrite it as a quotient, apply L'Hôpital's rule to find that the logarithm tends to 1, and exponentiate. The answer is $e^1 = e \approx 2.718281828$ . The same method shows that $x^x \to 1$ as $x \to 0^+$ .

Limits at infinity and horizontal asymptotes

When $x \to \pm\infty$ , rational functions are governed by their highest powers. If the degrees of the numerator and denominator are equal, the limit is the ratio of the leading coefficients, so $\frac{3x^2 + 1}{5x^2 - x} \to \frac35$ . If the denominator has the higher degree, the limit is 0; if the numerator does, the function grows without bound.

These limits are exactly the horizontal asymptotes of a graph. Exponentials eventually dominate any power and logarithms grow more slowly than any positive power, comparisons that L'Hôpital's rule makes precise.

When a limit does not exist

A two-sided limit fails to exist when the left and right limits differ, as with $\frac{|x|}{x}$ at 0, where the calculator reports a right-hand limit of 1 and a left-hand limit of $-1$ . It also fails when the function oscillates without settling, as $\sin\frac1x$ does near 0. In that situation the calculator explains that the values do not settle rather than inventing a number.

An infinite limit is technically another way of not existing as a real number, but it carries more information, so it is reported as $\infty$ or $-\infty$ with the side it applies to.

Common mistakes

Writing $\frac00 = 0$ or $\frac00 = 1$ ; the form is indeterminate and needs more work.
Applying L'Hôpital's rule to a limit that is not $\frac00$ or $\frac{\infty}{\infty}$ , such as $\frac{x+1}{x}$ at 0.
Differentiating the whole fraction with the quotient rule instead of differentiating the top and bottom separately.
Assuming $1^\infty = 1$ ; the limit of $(1 + \frac1x)^x$ is $e$ , not 1.
Reporting a two-sided limit when only one side has been checked.
Using degree mode on a calculator for trigonometric limits; $\frac{\sin x}{x} \to 1$ only in radians.

Tips

Always substitute first; many limits need nothing more.
Try factoring and cancelling before reaching for L'Hôpital's rule; it is often quicker.
For limits at infinity, divide every term by the highest power of x in the denominator.
Evaluate the function at a few points close to the target to estimate the answer before solving.
Rewrite powers with a variable in the exponent as $e^{g \ln f}$ .
Sketch or graph the function to see whether the two sides agree.

Frequently asked questions

What does the limit calculator do when substitution gives 0/0?