Limit Calculator
Compute the limit of a rational function (numerator_coeff · x^numerator_power) / (denominator_coeff · x^denominator_power) as x approaches a given value via direct substitution. Returns infinity when the denominator evaluates to zero.
Last updated: May 2026
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About this calculator
This calculator evaluates a simple rational expression of two power terms at a target x and returns the result, providing a direct-substitution limit estimate. The formula is L = (numerator_coeff · approach_value^numerator_power) / (denominator_coeff · approach_value^denominator_power), with a guard that returns infinity when the denominator equals zero. This is the easiest case of limits — when both numerator and denominator are continuous and the denominator is nonzero at the target, the limit equals the function value at that point. Edge cases: when denominator is zero at the target, the calculator returns infinity, but the actual mathematical answer depends on whether the numerator is also zero (indeterminate 0/0 form requiring L'Hôpital's rule or algebraic manipulation), nonzero (genuine vertical asymptote, ±∞ depending on sign), or whether the limit exists at all from one or both sides. The calculator cannot distinguish a 0/0 indeterminate form from a 'real' limit and will return NaN or undefined in that case. It also cannot handle limits at infinity (x → ∞), one-sided limits (x → c⁻ or x → c⁺), oscillating functions, removable discontinuities, or limits involving non-polynomial terms. Use only for direct substitution where the answer is genuinely the function value at the point.
How to use
Example 1 — straightforward substitution. limit of (3x²) / (2x) as x → 1. Inputs: numerator_coeff 3, numerator_power 2, denominator_coeff 2, denominator_power 1, approach_value 1. Step 1: numerator = 3 · 1² = 3. Step 2: denominator = 2 · 1¹ = 2. Step 3: result = 3 / 2 = 1.5. Verify: simplify (3x²)/(2x) = (3/2)x for x ≠ 0; at x = 1 this is 1.5 ✓. Note this is a removable discontinuity at x = 0 — the limit there is 0, not the calculator's infinity. Example 2 — limit at the origin with a vertical asymptote. limit of (5) / (2x²) as x → 0. Inputs: numerator_coeff 5, numerator_power 0, denominator_coeff 2, denominator_power 2, approach_value 0. Step 1: numerator = 5 · 0⁰ — note 0⁰ is conventionally 1 in this context, so = 5. Step 2: denominator = 2 · 0² = 0. Step 3: denominator is zero, calculator returns infinity. Verify: as x → 0, 5/(2x²) blows up since the denominator approaches zero while the numerator stays at 5; the two-sided limit is +∞ (both sides positive because of the squared denominator), so 'infinity' is the right answer ✓. For (5)/(2x) instead, the left and right limits are −∞ and +∞ respectively, so the two-sided limit does not exist — but the calculator would still output infinity, which is misleading.
Frequently asked questions
Why does the calculator return infinity for 0/0 cases?
It doesn't — it returns infinity only when the denominator is exactly zero. For 0/0, JavaScript's arithmetic produces NaN (0 divided by 0 is undefined), which the calculator may display as NaN or as some other error. Mathematically, 0/0 is an indeterminate form, meaning the limit could be anything — zero, infinity, a finite nonzero number, or might not exist — and the answer depends on the behaviour of the numerator and denominator near the target, not their values at the target. To resolve 0/0 you need algebraic manipulation (factoring, conjugate multiplication), L'Hôpital's rule (differentiate numerator and denominator separately and try again), or a series expansion. This calculator cannot do any of those — it does direct substitution only. If your inputs give 0/0, treat the result as 'use a different method' rather than as the actual limit.
What's the difference between the limit and the function value at a point?
The limit lim_{x→c} f(x) describes the value f(x) approaches as x gets arbitrarily close to c without necessarily equaling c. The function value f(c) is just what the function evaluates to at the exact point. For continuous functions these are equal, which is why direct substitution works — and this calculator implicitly assumes continuity. For functions with removable discontinuities (a 'hole' at c), the limit exists but the function value either doesn't or differs from the limit; classic example is (x²−1)/(x−1), which simplifies to x+1 everywhere except x = 1, has limit 2 at x = 1, but is undefined there. For functions with jumps, the left limit and right limit differ and the two-sided limit doesn't exist. For functions with vertical asymptotes, the limit is ±∞ rather than a finite number. This calculator can only handle the continuous case; for the others you need additional analysis.
How does this differ from L'Hôpital's rule?
Direct substitution (what this calculator does) works only when both numerator and denominator are continuous at the target and the denominator is nonzero — the easy case. L'Hôpital's rule handles the indeterminate forms 0/0 and ±∞/±∞ by differentiating numerator and denominator separately and taking the limit of the resulting quotient: lim f/g = lim f'/g' under those conditions. For example, lim_{x→0} sin(x)/x is 0/0 by direct substitution, but L'Hôpital gives cos(0)/1 = 1, the correct limit. This calculator cannot apply L'Hôpital's rule — you would need a symbolic differentiation tool or to apply the rule by hand. For polynomial ratios as x → ∞ (also an indeterminate form), the limit equals the ratio of leading coefficients if the degrees match, zero if numerator degree is lower, and ±∞ if higher; again this calculator cannot do that automatically because it doesn't handle x → ∞.
What are the common mistakes when computing limits?
The biggest mistake is treating direct substitution as universal — it only works for continuous functions at non-removable points. If substitution gives 0/0, ∞/∞, ∞ − ∞, 0·∞, 0⁰, ∞⁰, or 1^∞, you have an indeterminate form and need L'Hôpital, algebraic manipulation, or series expansion, not a calculator answer. The second mistake is treating one-sided behaviour as the limit when left and right limits differ; the two-sided limit only exists when both sides agree, and a calculator giving 'infinity' could mask the fact that one side goes to +∞ and the other to −∞. The third is forgetting the sign analysis near asymptotes: 1/x has limit −∞ from the left and +∞ from the right of zero, so the two-sided limit does not exist. People also confuse 'limit exists' with 'function defined' — these are independent. Finally, this calculator cannot handle non-polynomial functions (trig, exponential, log, piecewise), and using it for those will give wrong or undefined answers.
When should I not use this calculator?
Do not use it for indeterminate forms (0/0, ∞/∞), where direct substitution fails and you need L'Hôpital's rule, algebraic simplification, or series expansion to resolve. It is not appropriate for limits at infinity (x → ∞ or x → −∞), one-sided limits, limits of piecewise functions, or limits where the answer depends on the direction of approach. Do not use it for non-polynomial functions — trigonometric, exponential, logarithmic, or any function not of the form c · x^n divided by d · x^m. It cannot handle removable discontinuities correctly (it will return NaN or infinity where the actual limit is finite). Avoid it for functions with oscillatory behaviour near the target (like sin(1/x) as x → 0, whose limit does not exist) — the direct-substitution approach silently misses non-existence. For any nontrivial limit problem, use a symbolic engine (Wolfram Alpha, SymPy, Maxima, Mathematica) or work by hand using standard limit theorems.