Unit 1: Limits & Continuity

Limit Laws & Algebraic Techniques

Applying limit laws and algebraic techniques to evaluate limits.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Estimating Limits

Evaluating Limits Without a Table

A table of values can only ever estimate a limit — it never proves the exact value. Before reading on: for limₓ→3 (x² − 9)/(x − 3), which also gives 0/0 on direct substitution, what algebraic move would remove the problem entirely, the way it did for the rational function in the last lesson?

Limits obey algebraic laws: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on, whenever the individual limits exist. These laws let you evaluate a limit built from simpler pieces without a table — but a 0/0 form means direct substitution can't be used yet, and some algebraic rewriting has to happen first.

Worked Example — Resolving 0/0 by Factoring

Evaluate limₓ→3 (x² − 9)/(x − 3). Direct substitution gives 0/0. Factor the numerator: (x − 3)(x + 3)/(x − 3) = x + 3, valid for x ≠ 3 — which is exactly what the limit cares about, since a limit never depends on the value at x = 3 itself. So limₓ→3 (x² − 9)/(x − 3) = limₓ→3 (x + 3) = 3 + 3 = 6.

Worked Example — Resolving 0/0 by Rationalizing

Evaluate limₓ→0 (√(x + 4) − 2)/x. Direct substitution gives (√4 − 2)/0 = 0/0. Multiply numerator and denominator by the conjugate, √(x + 4) + 2: [(x + 4) − 4] / [x(√(x + 4) + 2)] = x / [x(√(x + 4) + 2)] = 1/(√(x + 4) + 2), valid for x ≠ 0. As x → 0, this becomes 1/(√4 + 2) = 1/4. Check numerically: at x = 0.001, (√4.001 − 2)/0.001 ≈ 0.24998, matching 1/4 = 0.25.

Worked Example — Applying Limit Laws to Given Values

Given limₓ→2 f(x) = 5 and limₓ→2 g(x) = −2, find limₓ→2 [3f(x) + f(x)g(x)]. Apply the sum, constant-multiple, and product laws directly: 3·limf(x) + limf(x)·limg(x) = 3(5) + (5)(−2) = 15 − 10 = 5. No table or graph is needed — just the limit laws applied to the given values.

Tip

Try direct substitution first, always. Only reach for factoring or rationalizing when substitution produces an indeterminate form like 0/0 — most limits don't need any algebraic rewriting at all.

Common Mistakes

Canceling the (x − 3) factor and then substituting x = 3 into the original, uncancelled expression.
After factoring and cancelling, substitute into the simplified expression (x + 3), not the original (x² − 9)/(x − 3) — the whole point of cancelling was to remove the division-by-zero problem.
Multiplying by the conjugate of the wrong part of the expression, or forgetting to multiply both numerator and denominator.
Rationalizing requires multiplying the entire fraction by (conjugate)/(conjugate) — a form of multiplying by 1 — so both the numerator and denominator change together and the expression's value is preserved.

Key Takeaways

Limit laws let sums, products, quotients, and constant multiples of limits be evaluated from the limits of their pieces.
A 0/0 result from direct substitution means the limit might still exist — factoring or rationalizing can remove the shared zero factor causing the indeterminate form.
Limits can be evaluated purely from given limit values, without ever knowing f or g's actual formulas.

Summary

Limit laws and algebraic techniques evaluate limits exactly. The next lesson uses limits to define exactly what it means for a function to be continuous — and what happens when it isn't.

Track your progress

← Estimating Limits Back to Unit 1: Limits & Continuity Defining Continuity→