What does this tool do?
This limit calculator follows the method of the Moroccan 2nd-year Bac programme. You type f(x), choose the point (a number a, from the left a⁻ or from the right a⁺, or +∞ or −∞), and it writes the solution like a teacher: substitution, the named indeterminate form, factoring, the conjugate, dominant terms, usual limits or comparative growth.
It covers the functions shared by the Mathematical Sciences and Experimental Sciences streams: polynomials, rational functions, square roots, ln, exponential, sin, cos and their compositions. Each exact result is compared with values of f near the point before it is shown.
When no method from the course applies, the tool shows only an estimate marked “≈ estimate, not a proof”, never an invented solution. To continue an exercise, go on to the function study or the derivative.
How to use it
- Type the function with x, for example (x²−4)/(x−2), sqrt(x²+1)−x, ln(x)/x or sin(3x)/x.
- Choose what x tends to: a number a (both sides, left a⁻ or right a⁺), +∞ or −∞.
- Read the limit, the indeterminate form met, the method and the detailed solution; if f has an asymptote there, it is stated.
Write 2x for 2 × x, x² or x^2 for a power, sqrt(x) or √x for a root, e^x or exp(x) for the exponential. The five examples under the input fill in the function and the point. The Share button copies a link with the same function, point and side.
Calculation method
Usual limits: sin x / x → 1 · (eˣ − 1)/x → 1 · ln(1 + x)/x → 1 · (1 − cos x)/x² → 1/2 (at 0)
Comparative growth: ln x / x → 0 · eˣ / x → +∞ (at +∞) · x ln x → 0 (at 0⁺) · x eˣ → 0 (at −∞)
Always start by replacing x with a. If f is defined at a (a polynomial, a fraction whose denominator does not vanish, a root of a positive number), the limit is f(a). Otherwise identify the form: 0/0 for a fraction whose two parts vanish, k/0 when only the denominator vanishes, ∞/∞ and ∞ − ∞ at infinity, 0 × ∞ for a product.
For 0/0, factor (x − a) out of polynomials or multiply by the conjugate when a root appears. At infinity, a polynomial behaves like its highest-degree term and a fraction like the ratio of the dominant terms. For k/0, study the sign of the denominator on the left and on the right: if the signs differ, there is no limit.
An infinite limit at a gives the vertical asymptote x = a; a finite limit l at ±∞ gives the horizontal asymptote y = l. The squeeze theorem handles functions like sin(x)/x at +∞, squeezed between −1/x and 1/x.
Worked example
Example 1: lim x→2 (x² − 4)/(x − 2)
- Replace x by 2: x² − 4 → 0 and x − 2 → 0. This is the indeterminate form 0/0.
- Factor: x² − 4 = (x − 2)(x + 2). For x ≠ 2, f(x) = x + 2.
- x + 2 is continuous at 2, so the limit is 2 + 2 = 4.
- Numeric check: f(2.001) = 4.001.
lim x→2 (x² − 4)/(x − 2) = 4 (form 0/0 resolved by factoring).
Another scenario: lim x→+∞ (√(x² + x) − x)
- At +∞, √(x² + x) → +∞ and −x → −∞: this is the indeterminate form ∞ − ∞.
- Multiply and divide by the conjugate √(x² + x) + x: this gives x/(√(x² + x) + x).
- The dominant terms give x/(x + x) = 1/2, so the limit is 1/2.
- Numeric check: for x = 10,000, f(x) ≈ 0.499988.
lim x→+∞ (√(x² + x) − x) = 1/2: the line y = 1/2 is a horizontal asymptote.
Understanding the results
Indeterminate form
An indeterminate form is not a limit: it is a way of writing where the rules on limits cannot conclude. Depending on the function, 0/0 can be 4, 0 or +∞. f(x) has to be rewritten, and the tool names the form before applying the method.
Left and right limits
The limit at a exists only if the left and right limits are equal. For 1/x at 0 they are −∞ and +∞: no limit, yet the line x = 0 is still a vertical asymptote.
Usual limits
sin(x)/x, (eˣ − 1)/x and ln(1 + x)/x tend to 1 at 0. For sin(3x)/x, write 3 × sin(3x)/(3x) with X = 3x tending to 0, so the limit is 3.
Comparative growth
At +∞, the exponential beats every power of x, and every power beats ln x. So eˣ/x² → +∞ and ln(x)/x → 0. This rule resolves the forms ∞/∞ and 0 × ∞ involving ln and exp.
Tips and limitations
- Replace x by a first: many limits stop right there.
- Write formulas with x; for a root use sqrt(…) or √.
- If the signs differ on each side of a point, choose “from the left”, then “from the right”.
- An estimate is not a proof: in an exam, the method has to be written out.
Mistakes to avoid
- Writing 0/0 = 0 or ∞ − ∞ = 0: these are indeterminate forms.
- Forgetting the sign of the denominator for k/0: 1/(x − 3) is +∞ to the right of 3 and −∞ to the left.
- Keeping every term of a polynomial instead of the dominant one when x goes to infinity.
- Forgetting the domain: ln(x) at 0 can only be taken from the right.
Frequently asked questions
How do I resolve a 0/0 indeterminate form?
For a fraction of polynomials, factor (x − a) out of the numerator and the denominator, then simplify. For (x² − 4)/(x − 2) at 2: x² − 4 = (x − 2)(x + 2), so f(x) = x + 2 and the limit is 4.
What is the limit of 1/(x − 3) at 3?
The form is k/0, so the sign of the denominator is needed. To the right of 3, x − 3 > 0 and the limit is +∞; to the left, x − 3 < 0 and the limit is −∞. The two sides differ, so there is no limit at 3.
What if the tool only gives an estimate?
No method from the course was recognised. The value shown comes from computing f near the point: it helps guess the limit, but it is not a proof to copy into an answer.
How do I find a limit at infinity?
Identify the form, then keep the dominant term: (3x² + 1)/(x² − 5) behaves like 3x²/x² = 3. For a difference of roots, use the conjugate.
What does 0⁺ mean?
It is a limit taken through positive values. For example x ln x → 0 as x tends to 0 through positive values, and 1/x tends to +∞ at 0⁺.
How do limits give asymptotes?
An infinite limit at a gives the vertical asymptote x = a; a finite limit l at +∞ or −∞ gives the horizontal asymptote y = l. The tool points it out.