Limit Calculator
Evaluate limits of functions numerically as x approaches a specific value.
Quick Examples
Limit Properties
A limit describes the value a function approaches as the input approaches a specific value, even if the function is not defined at that point.
Limit Laws
lim[f(x) · g(x)] = lim f(x) · lim g(x)
lim[f(x)/g(x)] = lim f(x) / lim g(x) (if lim g(x) ≠ 0)
Important Limits
| Limit | Value |
|---|---|
| lim(x→0) sin(x)/x | 1 |
| lim(x→0) (e^x - 1)/x | 1 |
| lim(x→0) (1+x)^(1/x) | e |
| lim(x→∞) (1 + 1/x)^x | e |
| lim(x→0) ln(1+x)/x | 1 |
L'Hopital's Rule
When lim gives 0/0 or ∞/∞ indeterminate forms:
Worked Example: Direct Substitution
Evaluate lim(x→3) (2x² − 5x + 1). Polynomials are continuous everywhere, so substitute directly: 2(9) − 5(3) + 1 = 18 − 15 + 1 = 4. Most limits that are not indeterminate forms are handled this way.
Worked Example: Factoring a 0/0 Form
Evaluate lim(x→2) (x² − 4)/(x − 2):
- Try substitution: (4 − 4)/(2 − 2) = 0/0 — indeterminate.
- Factor: (x − 2)(x + 2)/(x − 2).
- Cancel the common factor (valid since x ≠ 2): lim(x→2) (x + 2).
- Substitute: 2 + 2 = 4.
Note the original function is undefined at x = 2 (a hole), yet the limit exists — the limit describes behavior near the point, not at it.
Worked Example: Rationalizing
Evaluate lim(x→0) (√(x+4) − 2)/x. Direct substitution gives 0/0. Multiply by the conjugate (√(x+4) + 2)/(√(x+4) + 2): the numerator becomes (x + 4) − 4 = x, which cancels with the denominator's x, leaving lim(x→0) 1/(√(x+4) + 2) = 1/4 = 0.25.
Worked Example: L'Hôpital's Rule
Evaluate lim(x→0) sin(x)/x:
- Check the form: sin(0)/0 = 0/0 — indeterminate, so L'Hôpital's rule applies.
- Differentiate numerator and denominator separately (not the quotient rule): lim(x→0) cos(x)/1.
- Substitute: cos(0) = 1.
Worked Example: Limit at Infinity
Evaluate lim(x→∞) (3x² + 2x − 5)/(x² + 1). Divide numerator and denominator by the highest power, x²: the 2/x, 5/x², and 1/x² terms all vanish, leaving 3/1 = 3. The line y = 3 is a horizontal asymptote.
Worked Example: Squeeze Theorem
Evaluate lim(x→0) x²·sin(1/x). Since −1 ≤ sin(1/x) ≤ 1, we have −x² ≤ x²·sin(1/x) ≤ x². Both bounds approach 0, so the limit is 0 — squeezed from both sides.
One-Sided Limits: When a Limit Does Not Exist
A two-sided limit exists only if the left-hand and right-hand limits agree. For f(x) = |x|/x: from the right the value is 1, from the left it is −1, so lim(x→0) |x|/x does not exist.
Indeterminate Forms
| Form | Example | Typical approach |
|---|---|---|
| 0/0 | lim(x→2) (x²−4)/(x−2) | Factor and cancel, or L'Hôpital's rule |
| ∞/∞ | lim(x→∞) x²/eˣ | L'Hôpital's rule (repeated if needed) |
| 0 × ∞ | lim(x→∞) x·e^(−x) | Rewrite as a quotient |
| ∞ − ∞ | lim(x→∞) (x² − x) | Factor out the dominant term |
| 0⁰, 1^∞, ∞⁰ | lim(x→0⁺) xˣ | Take the logarithm first |
Limits at Infinity: Degree Rules
For a rational function, compare the degrees of numerator and denominator:
- Denominator degree higher: the limit is 0.
- Degrees equal: the limit is the ratio of the leading coefficients.
- Numerator degree higher: the function diverges to ±∞.
Limits and Continuity
A function f is continuous at a point a when three conditions hold: f(a) is defined, lim(x→a) f(x) exists, and the limit equals f(a). Discontinuities come in three flavors: a removable hole (limit exists but f(a) is missing or different), a jump (one-sided limits disagree), and an infinite discontinuity (a vertical asymptote).
Where Limits Appear
- Instantaneous velocity: average speed over shrinking time intervals tends to the derivative.
- Area under a curve: Riemann sums converge to the definite integral as the rectangles narrow.
- Compound interest: lim(n→∞) (1 + r/n)^n = e^r — the number e is born from a limit.
Common Mistakes
- Using L'Hôpital's rule on a form that is not 0/0 or ∞/∞ — try direct substitution first, always.
- Differentiating the whole fraction instead of numerator and denominator separately.
- Treating ∞ as a number: ∞ − ∞ is indeterminate, not 0.
- Assuming the limit equals f(a) at a discontinuity — a hole or jump does not change the surrounding trend.
- Skipping the one-sided check: if left and right disagree, the two-sided limit does not exist.
Types of Limits
| Type | Notation | Example |
|---|---|---|
| Two-sided | lim(x→a) f(x) | lim(x→3) (2x²−5x+1) = 4 |
| Left-hand | lim(x→a⁻) f(x) | lim(x→0⁻) |x|/x = −1 |
| Right-hand | lim(x→a⁺) f(x) | lim(x→0⁺) |x|/x = 1 |
| At infinity | lim(x→∞) f(x) | lim(x→∞) (3x²+2x−5)/(x²+1) = 3 |
| Infinite | lim(x→a) f(x) = ∞ | lim(x→0⁺) 1/x = ∞, lim(x→0⁻) 1/x = −∞ |
Infinite limits signal a vertical asymptote: the function grows without bound near the point.
Continuity vs Differentiability
Differentiability implies continuity: if a derivative exists at a point, the function must be continuous there. The converse is false — f(x) = |x| is continuous at 0 but has a sharp corner, so no derivative exists there. Checking continuity is therefore the first diagnostic step before differentiating.
How to Use This Calculator
- Direct/symbolic evaluation: enter the function and the point; the tool applies substitution, factoring, or L'Hôpital's rule as needed.
- Graph: visualize the function's behavior near the point to spot holes, jumps, and asymptotes.
- Table of values: approach the point from both sides numerically and watch the trend.
If the symbolic result and the graph disagree, trust neither blindly — check the one-sided behavior, since that is where most surprises live.
Evaluation Strategy: A Decision Flow
- Try direct substitution. If it works, you are done.
- Got 0/0 or ∞/∞? Try factoring, cancelling, or rationalizing first.
- Still indeterminate? Apply L'Hôpital's rule (differentiate top and bottom separately).
- Repeated failure? Try the squeeze theorem, logarithms, or known special limits like lim(x→0) sin(x)/x = 1.
- One-sided disagreement? Then the two-sided limit does not exist — state which side does what.
Limits in Sequences
Limits also govern sequences: lim(n→∞) (1 + 1/n)ⁿ = e ≈ 2.71828. The same idea — behavior as the input grows without bound — connects discrete sequences to continuous functions, and it is how e appears in compound interest, population models, and radioactive decay.
When the Limit Equals the Function Value
For continuous functions the limit is just the function value: lim(x→2) eˣ = e² ≈ 7.389, lim(x→π/4) tan(x) = 1. The interesting limits are the ones where substitution fails — holes, jumps, asymptotes, and infinity. If substitution works, there is nothing left to prove.
Removable vs Essential Discontinuities
A hole (removable discontinuity) like (x² − 4)/(x − 2) at x = 2 has a limit — the function can be "repaired" by defining f(2) = 4. A jump or asymptote (essential discontinuity) cannot be repaired with a single value; the limit either disagrees from the sides or does not exist at all.