The Limit Laws

The limit laws — sum, difference, product, quotient, constant-multiple, and power — build limits of complicated expressions out of simpler ones. Their payoff: when a function is continuous at \(a\), you find the limit by plugging in: \(\lim_{x \to a} f(x) = f(a)\).

By the end you'll be able to evaluate limits of polynomials and rational functions by direct substitution, and spot the one condition — a denominator whose limit is 0 — where the quotient law refuses to help.

Predict: which limit law stops working as a slides toward −4, where the denominator heads to 0? Slide and watch the readout.

This is \(g(x) = \dfrac{3x^2 - 5x + 1}{x + 4}\), a rational function. Drag a and the point rides the curve while the readout evaluates \(g(a)\) by substitution. The denominator value a + 4 is color-linked to the dashed guide line at x = −4 — the one x where substitution breaks. Hover or focus the point for exact values.

a = 2.0 · denominator a + 4 = 6.0 · g(a) = 0.500 — g is continuous here, so limit = g(a)

Approaching the break at x = −4 — the denominator is closing in on 0, where the quotient law's substitution shortcut will stop applying.

g(x) = (3x² − 5x + 1)/(x + 4) — the point evaluates g by substitution; the dashed line marks x = −4, where the denominator's limit is 0
g(x) point at (a, g(a)) x = −4 (denominator → 0)

The limit laws let you evaluate the limit of any polynomial or rational expression piece by piece — and for a function continuous at \(a\), the whole computation collapses to one step: plug in \(a\).

Formal

Suppose \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\) both exist. Then the limit of a combination is the same combination of the limits:

  • Sum / Difference: \(\lim\, [f \pm g] = L \pm M\)
  • Constant multiple: \(\lim\, [c \cdot f] = c\,L\)
  • Product: \(\lim\, [f \cdot g] = L \cdot M\)
  • Quotient: \(\lim\, [f / g] = L / M\), provided \(M \neq 0\) — the same condition the guide line flags in the demo above
  • Power / Root: \(\lim\, [f]^n = L^n\), and \(\lim \sqrt[n]{f} = \sqrt[n]{L}\) (root valid when defined)

Together with the two atoms \(\lim_{x \to a} c = c\) and \(\lim_{x \to a} x = a\), these laws assemble the limit of any polynomial or rational expression from its pieces. The payoff: whenever \(f\) is continuous at \(a\), \(\lim_{x \to a} f(x) = f(a)\).

Applied

In practice you almost never build a limit law by law. Every polynomial is continuous everywhere, and a rational function is continuous everywhere its denominator is nonzero — so for these functions, "take the limit" simply means "evaluate at \(a\)." Your working procedure: check the denominator at \(a\). If it's nonzero, substitute and you're done — that's the moving point in the demo reading off \(g(a)\). If the denominator's limit IS 0, stop: substitution gives \(0/0\) or a blow-up, the quotient law doesn't apply, and you need the algebraic tools of the next lesson.

Worked example

Evaluate \(\lim_{x \to -1} \dfrac{3x^2 - 5x + 1}{x + 4}\). Check the denominator at \(-1\): it's \(-1 + 4 = 3 \neq 0\), so \(g\) is continuous there and direct substitution is valid. Numerator: \(3(-1)^2 - 5(-1) + 1 = 3 + 5 + 1 = 9\). Denominator: \(3\). So the limit is \(9/3 = \) \(3\).

Your turn

Same function, a different target: \(\lim_{x \to 1} \dfrac{3x^2 - 5x + 1}{x + 4}\). The denominator at \(1\) is \(1 + 4 = 5 \neq 0\), so substitution is valid. Numerator: \(3(1)^2 - 5(1) + 1 = 3 - 5 + 1 = -1\). Now finish it: the limit is ____

Reveal the answer

Limit \(= -1/5 = \) \(-0.2\). Set \(a = 1\) on the slider above and check it against the readout — the point sits at \((1, -0.2)\), just below the x-axis.

More info — why "plug it in" is a theorem, not a shortcut

Rebuild \(\lim_{x \to 2} (3x^2 - 5x + 1)\) from the two atoms: \(\lim_{x \to 2} x = 2\), so the product and power laws give \(\lim x^2 = 4\); the constant-multiple law gives \(\lim 3x^2 = 12\) and \(\lim 5x = 10\); and the sum/difference laws combine them: \(12 - 10 + 1 = 3\). Notice that the answer is exactly the polynomial evaluated at 2 — every step of the law-by-law build-up mimics the arithmetic of substitution. That's why "limit = \(f(a)\)" holds for every polynomial, and (adding the quotient law) for every rational function where the denominator is nonzero. From the previous lesson, remember what makes this special: in general a limit is about values NEAR \(a\) and can differ from \(f(a)\) — continuity is precisely the case where the two agree. The OpenStax chapter in Dive deeper below proves each law from the limit definition.

Check your understanding

Question 1 of 4

Evaluate \(\lim_{x \to 0} \dfrac{3x^2 - 5x + 1}{x + 4}\).

Question 2 of 4

Suppose \(\lim_{x \to a} f(x) = 5\) and \(\lim_{x \to a} g(x) = 0\). Which limit can you NOT evaluate using the limit laws alone?

Question 3 of 4

True or false: the quotient law gives \(\lim_{x \to 1} \dfrac{x-1}{x^2-1} = \frac{0}{0} = 0\).

Question 4 of 4

A function \(h\) satisfies \(\lim_{x \to 3} h(x) = 7\) but \(h(3) = 2\). What can you conclude?

Recap

  • If \(\lim f = L\) and \(\lim g = M\) exist, then limits pass through sums, differences, constant multiples, products, powers/roots — and quotients when \(M \neq 0\).
  • Atoms: \(\lim_{x \to a} c = c\) and \(\lim_{x \to a} x = a\) — the laws build everything else from these.
  • If \(f\) is continuous at \(a\) — every polynomial everywhere, every rational function where its denominator is nonzero — then \(\lim_{x \to a} f(x) = f(a)\): just substitute.
  • The quotient law does NOT apply when the denominator's limit is 0 — substitution gives \(0/0\) or a blow-up, and you must resolve it another way.

Dive deeper

Sources

  • The Limit Laws and Direct Substitution