Algebra of Limits and Polynomial Limits

The Limit Laws

Limits respect arithmetic. If $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist, then sums, differences, products and quotients behave exactly as you hope:

Key Point (Theorem 1): $\lim [f \pm g] = \lim f \pm \lim g$; $\lim [f \cdot g] = \lim f \cdot \lim g$; $\lim \frac{f}{g} = \frac{\lim f}{\lim g}$ when $\lim g \ne 0$; and $\lim [\lambda f] = \lambda \lim f$ for a constant $\lambda$.

Polynomials: just substitute. Since $\lim_{x \to a} x = a$, the product law gives $\lim_{x \to a} x^n = a^n$, and the sum law assembles any polynomial:

$$\lim_{x \to a} f(x) = f(a) \quad \text{for every polynomial } f$$

Rational functions $\frac{g(x)}{h(x)}$: substitute first. If $h(a) \ne 0$ the answer is $\frac{g(a)}{h(a)}$. If $h(a) = 0$ but $g(a) \ne 0$, the limit does not exist. If both vanish — the $\frac{0}{0}$ form — factor the vanishing power of $(x - a)$ out of top and bottom and cancel:

Cancelling $(x - a)$ is legal precisely because $x \to a$ means $x \ne a$ throughout the approach.

The Workhorse: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$

Key Point (Theorem 2): for any positive integer $n$ — and in fact any rational $n$ (with $a > 0$ where needed) —

$$\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$$

Why: $x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + \cdots + a^{n-1})$; after cancelling, each of the $n$ surviving terms tends to $a^{n-1}$.

Using it: $\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$ — divide top and bottom by $x - 1$ and apply the theorem twice: $\frac{15}{10} = \frac{3}{2}$. And $\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$: substitute $y = 1 + x$ to get $\lim_{y \to 1} \frac{y^{1/2} - 1}{y - 1} = \frac{1}{2}$ — the rational-exponent case at work.

Recognition drill: any limit of the shape $\frac{(\text{power}) - (\text{same power at } a)}{x - a}$ is this theorem; so is $\frac{(x+1)^5 - 1}{x}$ after the shift $y = x + 1$.

[Board Tip] Quote the theorem by name-and-statement when you use it — "by $\lim_{x \to a}\frac{x^n - a^n}{x - a} = na^{n-1}$" — the step earns its mark only when the examiner can see which tool you applied.

Solved Examples

Example 1: Polynomials by substitution

Find: (i) $\lim_{x \to 1}[x^3 - x^2 + 1]$; (ii) $\lim_{x \to 3}[x(x + 1)]$; (iii) $\lim_{x \to -1}[1 + x + x^2 + \cdots + x^{10}]$.

Solution:

Step 1 — (i) Substitute. Polynomials obey $\lim =$ value: $1 - 1 + 1 = 1$.

Step 2 — (ii). $3 \times (3 + 1) = 12$.

Step 3 — (iii) Count the alternation. At $x = -1$ the eleven terms run $1 - 1 + 1 - \cdots + 1$: five cancelling pairs plus a final $+1$ — the limit is 1.

Takeaway: For any polynomial, the limit is one substitution; the only care needed is arithmetic, as in the alternating sum.

Example 2: Direct substitution

Evaluate: (i) $\lim_{x \to 3}(x + 3)$; (ii) $\lim_{x \to \pi}\left(x - \frac{22}{7}\right)$; (iii) $\lim_{r \to 1} \pi r^2$.

Solution:

Step 1 — (i). $3 + 3 = 6$.

Step 2 — (ii). $\pi - \frac{22}{7}$ — and this is NOT zero: $\frac{22}{7} = 3.142857\ldots$ is only an approximation of $\pi = 3.14159\ldots$; the limit is a small negative number.

Step 3 — (iii). $\pi(1)^2 = \pi$.

Takeaway: Part (ii) is a designed trap — $\pi \ne \frac{22}{7}$; leave the answer symbolic.

Example 3: Rational, denominator alive

Evaluate: (i) $\lim_{x \to 4} \frac{4x + 3}{x - 2}$; (ii) $\lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1}$.

Solution:

Step 1 — Check the denominator FIRST. (i) At 4: $x - 2 = 2 \ne 0$ — safe to substitute.

Step 2 — (i) Substitute. $\frac{16 + 3}{2} = \frac{19}{2}$.

Step 3 — (ii) Same check, then substitute. Denominator $-2 \ne 0$: $\frac{1 - 1 + 1}{-2} = -\frac{1}{2}$.

Takeaway: A rational limit begins with one question — is the denominator non-zero at the point? If yes, it is a substitution problem.

Example 4: The 0/0 factor-cancel

Evaluate: (i) $\lim_{x \to 2} \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$; (ii) $\lim_{x \to 2} \frac{x^2 - 4}{x^3 - 4x^2 + 4x}$; (iii) $\lim_{x \to 2} \frac{x^3 - 2x^2}{x^2 - 5x + 6}$.

Solution:

Step 1 — (i) Factor out the vanishing power. $\frac{x(x-2)^2}{(x+2)(x-2)} = \frac{x(x-2)}{x+2}$; substituting, $\frac{2 \times 0}{4} = 0$.

Step 2 — (ii) The reciprocal flips the story. $\frac{(x+2)(x-2)}{x(x-2)^2} = \frac{x+2}{x(x-2)} \to \frac{4}{0}$: numerator alive, denominator dead — the limit does not exist.

Step 3 — (iii) Factor and cancel once. $\frac{x^2(x-2)}{(x-2)(x-3)} = \frac{x^2}{x-3} \to \frac{4}{-1} = -4$.

Takeaway: After cancelling, re-substitute and READ the form: $\frac{0}{4}$ is 0, $\frac{4}{0}$ is no limit, $\frac{4}{-1}$ is just arithmetic.

Example 5: Combine first, then cancel

Evaluate $\lim_{x \to 1}\left[\frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x}\right]$.

Solution:

Step 1 — Factor the denominators. $x^2 - x = x(x-1)$; $x^3 - 3x^2 + 2x = x(x-1)(x-2)$.

Step 2 — Combine over the common denominator $x(x-1)(x-2)$. Numerator: $(x-2)^2 - 1 = x^2 - 4x + 3 = (x-3)(x-1)$.

Step 3 — Cancel and substitute. $\frac{(x-3)(x-1)}{x(x-1)(x-2)} = \frac{x-3}{x(x-2)} \to \frac{-2}{1 \times (-1)} = 2$.

Takeaway: When each piece separately blows up, combine into ONE fraction first — the infinities cancel on paper before they can mislead.

Example 6: Shifted powers and quadratics

Evaluate: (i) $\lim_{x \to 0} \frac{(x+1)^5 - 1}{x}$; (ii) $\lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4}$.

Solution:

Step 1 — (i) Shift into workhorse form. Put $y = x + 1$ (so $y \to 1$): $\lim_{y \to 1} \frac{y^5 - 1}{y - 1}$.

Step 2 — (i) Apply the theorem. $5(1)^4 = 5$.

Step 3 — (ii) Factor the 0/0. $\frac{(x-2)(3x+5)}{(x-2)(x+2)} = \frac{3x+5}{x+2} \to \frac{11}{4}$.

Takeaway: $(x + 1)^n - 1$ over $x$ is the workhorse in disguise — one substitution reveals it.

Example 7: Higher-degree factoring

Evaluate $\lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3}$.

Solution:

Step 1 — Factor top and bottom. $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x-3)(x+3)(x^2+9)$; $2x^2 - 5x - 3 = (x-3)(2x+1)$.

Step 2 — Cancel $(x - 3)$. Left with $\frac{(x+3)(x^2+9)}{2x+1}$.

Step 3 — Substitute. $\frac{6 \times 18}{7} = \frac{108}{7}$.

Takeaway: Difference of squares twice ($x^4 - 81 \to x^2 - 9 \to x - 3$) — factor DOWN to the vanishing linear factor, no further.

Example 8: Parameters ride along

Evaluate: (i) $\lim_{x \to 0} \frac{ax + b}{cx + 1}$; (ii) $\lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}$ given $a + b + c \ne 0$.

Solution:

Step 1 — (i) Denominator at 0. $c(0) + 1 = 1 \ne 0$: substitute.

Step 2 — (i) Read off. $\frac{b}{1} = b$.

Step 3 — (ii) Substitute at 1. Top: $a + b + c$; bottom: $c + b + a$ — identical sums. The limit is 1, and the hypothesis $a + b + c \ne 0$ is exactly what keeps $\frac{0}{0}$ away.

Takeaway: Letters obey the same laws as numbers — check the denominator, substitute, simplify.

Example 9: Fractional powers via substitution

Evaluate $\lim_{z \to 1} \frac{z^{1/3} - 1}{z^{1/6} - 1}$.

Solution:

Step 1 — Clear the fractions with one substitution. Let $z = t^6$ (as $z \to 1$, $t \to 1$): $z^{1/3} = t^2$ and $z^{1/6} = t$.

Step 2 — Simplify. $\frac{t^2 - 1}{t - 1} = t + 1$ for $t \ne 1$.

Step 3 — Conclude. The limit is $1 + 1 = 2$. (Or apply the workhorse twice: $\frac{1/3}{1/6} = 2$ ✓.)

Takeaway: For mixed fractional exponents substitute $z = t^k$ with $k$ the LCM of the denominators — every power turns into an integer power.

Example 10: Combine, then cancel

Evaluate $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$.

Solution:

Step 1 — Combine the numerator. $\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$.

Step 2 — Divide. $\frac{x + 2}{2x(x + 2)} = \frac{1}{2x}$ for $x \ne -2$.

Step 3 — Substitute. $\frac{1}{2(-2)} = -\frac{1}{4}$.

Takeaway: The $(x + 2)$ upstairs was hiding inside the little fractions — simplify complex fractions before touching the limit.

Example 11: The workhorse twice

Evaluate: (i) $\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$; (ii) $\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$.

Solution:

Step 1 — (i) Divide top and bottom by $x - 1$. $\frac{(x^{15}-1)/(x-1)}{(x^{10}-1)/(x-1)} \to \frac{15(1)^{14}}{10(1)^9}$.

Step 2 — (i) Simplify. $\frac{15}{10} = \frac{3}{2}$.

Step 3 — (ii) Shift to base 1. With $y = 1 + x$: $\lim_{y \to 1} \frac{y^{1/2} - 1}{y - 1} = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}$.

Takeaway: Splitting one ratio into two workhorse quotients (divide both by $x - a$) is the standard trick for $\frac{x^m - a^m}{x^n - a^n}$: answer $\frac{m}{n}a^{m-n}$.

Example 12: Workhorse with a general exponent

If $\lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80$ for a positive integer $n$, find $n$.

Solution:

Step 1 — Apply the theorem. The limit equals $n \cdot 2^{n-1}$.

Step 2 — Solve $n \cdot 2^{n-1} = 80$. Test small $n$: $n = 4$: $4 \times 8 = 32$; $n = 5$: $5 \times 16 = 80$ ✓.

Step 3 — Conclude. $n = 5$ (and the function $n \cdot 2^{n-1}$ is increasing, so the solution is unique).

Takeaway: Parameter versions of the workhorse reduce to small integer equations — test values rather than solving formally.