Limit Calculator (Evaluate Limits Step by Step)
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To find the limit of (x² − 9)/(x − 3) as x → 3, factor the top as (x − 3)(x + 3), cancel the common factor x − 3, then substitute: 3 + 3 = 6. The limit calculator below does this for any function, from either side or at ±∞, and names the technique it used at every step.
Open circle: the point the graph approaches. It does not matter whether f is defined there.
| h | f(3 − h), left | f(3 + h), right |
|---|---|---|
| 0.1 | 5.9 | 6.1 |
| 0.01 | 5.99 | 6.01 |
| 0.001 | 5.999 | 6.001 |
| 0.0001 | 5.9999 | 6.0001 |
| 0.00001 | 5.99999 | 6.00001 |
Show steps (5)
- Indeterminate form 0/0Plugging in x = 3 gives 0/0. That form does not decide the limit, so rewrite the expression first.
- Factor and cancelx2 − 9x − 3 = (x − 3)(x + 3)(x − 3)Cancel the common factor (x − 3). This is allowed because x only approaches 3 and never equals it, so the factor is not zero. = x + 3
- Substitute x = 3The new expression is continuous at x = 3, so plug the value in: x + 3 → 3 + 3 = 6
- Conclusionlim(x→3) f(x) = 6.
- Check with numbersf(3 + 0.0001) = 6.0001; f(3 − 0.0001) = 5.9999. Both agree with the limit 6.
Next: differentiate f · limit definition of the derivative · graph it
How to use the limit calculator
- Type the function, for example (x^2 – 9)/(x – 3), sin(x)/x or (1 + 1/x)^x. The “Reads as” line shows the limit as it was understood, so a missing bracket is easy to spot.
- Enter the point x approaches: a number such as 3 or pi/2, or inf and -inf for limits at infinity.
- Choose the direction: both sides, from the left (x → a⁻) or from the right (x → a⁺). For a two-sided limit the tool works out each side separately and compares them.
- Read the answer, then open Show steps. Each step names the technique: direct substitution, factoring and canceling, the conjugate, standard trig limits, dominant terms, L’Hôpital’s rule, the squeeze theorem or a log rewrite.
- Use the graph and the table of values to see the answer. When the limit does not exist, the result says exactly why.
What a limit means
The limit of f(x) as x approaches a is the value f(x) gets close to when x gets close to a, without x ever being equal to a. The value at the point itself does not matter. For f(x) = (x² − 1)/(x − 1), the function is undefined at x = 1, but for every other x it equals x + 1, so the limit as x → 1 is 2.
A one-sided limit looks from one side only. The left-hand limit uses x < a and is written x → a⁻; the right-hand limit uses x > a and is written x → a⁺. The two-sided limit exists only when both one-sided limits exist and are equal.
A limit at infinity asks what happens to f(x) as x grows without bound. If the answer is a number L, the line y = L is a horizontal asymptote of the graph.
Limit techniques, in the order to try them
| Technique | Use it when | Example | Limit |
|---|---|---|---|
| Direct substitution | f is continuous at a (no division by 0, no root of a negative) | x² + 1 as x → 2 | 5 |
| Factor and cancel | 0/0 with polynomials | (x³ − 8)/(x − 2) as x → 2 | 12 |
| Multiply by the conjugate | 0/0 or ∞ − ∞ with a square root | (√(x + 4) − 2)/x as x → 0 | 1/4 |
| Standard limits | sin, tan, 1 − cos, eᵘ − 1 or ln(1 + u) with u → 0 | sin(3x)/sin(5x) as x → 0 | 3/5 |
| Dominant terms | rational or root expressions as x → ±∞ | (3x² + 1)/(2x² − x) as x → ∞ | 3/2 |
| L’Hôpital’s rule | 0/0 or ∞/∞ when algebra gets stuck | (eˣ − 1 − x)/x² as x → 0 | 1/2 |
| Squeeze theorem | a bounded factor times something going to 0 | x² sin(1/x) as x → 0 | 0 |
| Take the log | forms 1^∞, 0⁰, ∞⁰ | (1 + 1/x)ˣ as x → ∞ | e |
Always try substitution first. Only the forms 0/0, ∞/∞, 0·∞, ∞ − ∞, 1^∞, 0⁰ and ∞⁰ are indeterminate, meaning they do not decide the limit on their own. A nonzero number over 0, such as 5/0, is not indeterminate: it signals an infinite limit or a limit that does not exist.
Worked examples
Factoring: (x³ − 8)/(x − 2) as x → 2
Substitution gives 0/0. Factor the difference of cubes: x³ − 8 = (x − 2)(x² + 2x + 4). Cancel x − 2, which is never zero while x only approaches 2. Now substitute: 2² + 2·2 + 4 = 12.
Conjugate: (√(x + 4) − 2)/x as x → 0
Substitution gives 0/0. Multiply top and bottom by √(x + 4) + 2. The top becomes (x + 4) − 4 = x, so the expression is x / [x(√(x + 4) + 2)] = 1/(√(x + 4) + 2). Substitute x = 0: 1/(2 + 2) = 1/4.
At infinity: (3x² + 1)/(2x² − x) as x → ∞
For large x only the highest powers matter: the top behaves like 3x² and the bottom like 2x², so the quotient approaches 3/2. Dividing every term by x² gives the same result: (3 + 1/x²)/(2 − 1/x) → 3/2. When the top has the higher power the limit is ±∞; when the bottom does, it is 0.
L’Hôpital’s rule: (eˣ − 1 − x)/x² as x → 0
Substitution gives 0/0, so differentiate the top and the bottom separately: (eˣ − 1)/(2x). That is still 0/0, so apply the rule again: eˣ/2 → 1/2. L’Hôpital’s rule applies only to 0/0 and ∞/∞; using it on any other form gives wrong answers.
When a limit does not exist
“Does not exist” is a real answer, and it always has a reason:
- The sides disagree (jump). |x − 2|/(x − 2) equals −1 to the left of 2 and 1 to the right, so there is no two-sided limit.
- The values blow up. 1/x goes to −∞ from the left of 0 and to +∞ from the right. For 1/x² both sides go to +∞, and we write the limit as ∞, which still means it is not a finite number.
- The values oscillate. sin(1/x) swings between −1 and 1 infinitely often near 0 and never settles.
- The function is not defined on one side. √x has no real values for x < 0, so only the right-hand limit (0) exists. Some textbooks still write the limit as 0 because the domain stops at 0; the calculator shows both facts.
Estimating a limit with a table of values
A table is the quickest check. To estimate a limit as x → 3, evaluate f at 2.9, 2.99, 2.999 and at 3.1, 3.01, 3.001. For (x² − 9)/(x − 3) the left column reads 5.9, 5.99, 5.999 and the right column 6.1, 6.01, 6.001, so both sides point to 6. A table suggests the answer but does not prove it, and very small steps can suffer from rounding, so use it to check the algebra rather than replace it. The calculator prints this table for every problem, with h from 0.1 down to 0.00001.
Common mistakes
| Mistake | Fix |
|---|---|
| Writing 0/0 = 1 or 0/0 = 0 | 0/0 is a signal to rewrite, not an answer |
| Using L’Hôpital’s rule on 5/0 or 1/∞ | Check the form first; the rule needs 0/0 or ∞/∞ |
| Applying the quotient rule in L’Hôpital | Differentiate the top and the bottom separately |
| Reporting a two-sided limit from one side | Check both sides, especially with |x|, roots and fractions |
| Forgetting that √(x²) = |x| | As x → −∞, √(x² + 1) behaves like −x, not x |
| Typing sin x/x without brackets | Type sin(x)/x and read the “Reads as” line |
Using a limit calculator for homework
Work the problem by hand first: try substitution, name the form, choose a technique, and finish with substitution. Then enter the same limit here and compare each step. If your answer differs, the steps show where: a sign error when factoring, a missed side, or a rule used on the wrong form.
Quick check: the answer should match the table of values from both sides and the open circle on the graph.
Limits practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
Factor: (x − 4)(x + 4)/(x − 4) = x + 4 for x ≠ 4. Substitute 4: 4 + 4 = 8. The 0/0 form is a signal to simplify, not the answer.
Use sin u/u → 1 with u = 4x: sin(4x)/x = 4 · sin(4x)/(4x) → 4 · 1 = 4.
Equal highest powers, so the limit is the ratio of the leading coefficients: 5/2.
The bottom has the higher power (x² against x), so the quotient shrinks to 0.
Multiply by the conjugate √x + 2: (x − 4)/[(x − 4)(√x + 2)] = 1/(√x + 2) → 1/(2 + 2) = 1/4.
For x just above 2, x − 2 is a small positive number, so 1/(x − 2) is large and positive.
A function can be undefined at the point and still have a limit. Here the two sides disagree: −x/x = −1 on the left and x/x = 1 on the right.
Use the standard limit (1 − cos u)/u² → 1/2, or apply L’Hôpital’s rule twice: sin x/(2x) → cos x/2 → 1/2.
A nonzero number over 0 means the values blow up (an infinite limit or no limit). The other three forms need more work to decide.
The form is ∞/∞. L’Hôpital’s rule gives 1/eˣ, which goes to 0 because eˣ grows without bound.
Quiz complete
Frequently asked questions
What is a limit in calculus?
A limit is the value a function approaches as the input approaches a point. For (x² − 1)/(x − 1), the function is undefined at x = 1, but it equals x + 1 everywhere else, so the limit as x → 1 is 2. Limits describe behavior near a point, not the value at it.
How do you find a limit with a calculator?
Evaluate the function closer and closer to the point from both sides, or graph it and trace toward the point. For sin(x)/x as x → 0, values such as 0.998 at x = 0.1 and 0.99998 at x = 0.01 point to a limit of 1. A symbolic tool like the one above gives the exact value and the steps.
What does lim mean?
"lim" is short for limit. The notation lim(x → 3) (x² − 9)/(x − 3) = 6 reads "the limit of (x² − 9)/(x − 3) as x approaches 3 is 6." A small ⁺ or ⁻ after the point means the limit is taken from the right or from the left.
Are limits part of calculus?
Yes. Limits are the first major topic of Calculus 1, and both the derivative and the definite integral are defined as limits. The derivative, for example, is the limit of the difference quotient [f(x + h) − f(x)]/h as h → 0.
Does limit mean minimum or maximum?
No. A limit is the value the function approaches, which need not be the largest or smallest value. For example, sin(x)/x approaches 0 as x → ∞ while it keeps crossing above and below 0.
Can a limit be infinite?
The values can grow without bound, as with 1/x² near 0. We write the limit as ∞ to describe that behavior, but the limit does not exist as a real number, and the graph has a vertical asymptote there.
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