Limits of algebraic functions

Once you know what a limit means, the next question is purely mechanical: how do you actually find one? For the algebraic functions of Class 11 — polynomials and rational (ratio-of-polynomial) functions — the answer starts simple. Direct substitution works whenever the denominator does not vanish. The interesting cases are the ones where substitution jams on the form $\tfrac{0}{0}$ , a signal that a hidden common factor is hiding in plain sight. This article turns that signal into a routine: factorise or rationalise to expose and cancel the troublesome factor, then substitute. Both routes are really the same trick in different clothing, and both are packaged into a single standard formula, $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ , that will resurface almost immediately when differentiation by first principles is introduced.

The core insight

Limits of polynomials and rational functions are found by direct substitution wherever the denominator is nonzero; where substitution gives $\tfrac{0}{0}$ , factorising (or rationalising, for surds) exposes and cancels the common factor causing the zero, after which substitution works.

The core idea

Polynomials substitute directly. For $p(x) = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n$ , the sum and product rules for limits guarantee that the limit of the whole expression equals the expression evaluated at $a$ :

\lim_{x\to a} p(x) = p(a)

where: $p(x)$ is any polynomial and $a$ is the point being approached. No factor ever needs cancelling here, because a polynomial can never produce an indeterminate form on its own.

Rational functions substitute directly too — until the denominator vanishes. For $f(x) = \dfrac{p(x)}{q(x)}$ , the quotient rule gives

\lim_{x\to a}\frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}, \qquad \text{provided } q(a)\neq 0

where: $p(x)$ and $q(x)$ are polynomials, and the proviso $q(a) \ne 0$ is essential — division by zero is never permitted, not even inside a limit.

When $q(a) = 0$ and $p(a) = 0$ too, substitution collapses to $\tfrac{0}{0}$ . This is not "the answer is zero" and not "the limit fails to exist" — it is an indeterminate form, meaning the substitution step has given no information at all, and a different method is needed before any conclusion can be drawn. The fix is always the same idea: find and cancel the shared factor that is forcing both top and bottom to zero at $x=a$ , so the resulting simplified expression no longer has that problem at $x=a$ . Two techniques expose that shared factor. Factorisation works when $p(x)$ and $q(x)$ are ordinary polynomials — write each as a product and cancel the common linear factor $(x-a)$ . Rationalisation works when a surd (square root) sits in the numerator or denominator — multiply top and bottom by the conjugate so the surd disappears, turning it into the same kind of polynomial factor that factorisation would have produced. Both techniques rely on one quiet justification: cancelling $(x-a)$ is legitimate only because $x \to a$ means $x$ gets arbitrarily close to $a$ without ever equalling it, so $x - a \ne 0$ throughout the process even though $a$ is the target. The single most useful packaged result for this chapter is the standard algebraic limit

\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}, \qquad n \text{ rational}, \ a>0 \text{ when } n \text{ is not an integer}

where: $n$ is the exponent (an integer for the polynomial case, or a fraction such as $\tfrac12$ when a surd is involved), and $a$ is the nonzero point of approach. This single formula quietly covers both the factorisation case and the rationalisation case, because for a positive integer $n$ the factorisation $x^n - a^n = (x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$ cancels $(x-a)$ exactly, while for fractional $n$ (surds) the same limiting value emerges once the conjugate trick is carried out.

Figure

A flowchart for evaluating a limit of an algebraic function starting from substitution, branching to a direct answer when the result is defined and to factorise or rationalise then cancel then substitute when the result is zero over zero

Derivation AThe standard limit x to the n minus a to the n over x minus a

The goal is to show, for a positive integer $n$ , that $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$ by exposing the common factor $(x-a)$ through factorisation and cancelling it, since direct substitution at $x=a$ gives the indeterminate form $\tfrac{0}{0}$ and therefore carries no information on its own.

Step 1. Confirm the indeterminate form. At $x=a$ , the numerator is $a^n - a^n = 0$ and the denominator is $a - a = 0$ , so substitution gives $\tfrac{0}{0}$ — direct substitution cannot be used, and a different route is required.

Step 2. Factorise the numerator using the standard algebraic identity $x^n - a^n = (x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+\cdots+xa^{n-2}+a^{n-1})$ . This identity can be checked by expanding the right side: every cross term except $x^n$ and $-a^n$ cancels in pairs, leaving exactly $x^n - a^n$ .

Step 3. Cancel the common factor $(x-a)$ from numerator and denominator. This step is valid precisely because $x \to a$ means $x$ is never actually equal to $a$ during the limiting process, so $x - a \ne 0$ throughout:

\frac{x^n-a^n}{x-a} = \frac{(x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})}{x-a} = x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}

Step 4. Substitute $x=a$ into the simplified (now polynomial, no longer indeterminate) expression. There are exactly $n$ terms in the sum, and every single one becomes $a^{n-1}$ once $x=a$ :

\lim_{x\to a}\left(x^{n-1}+x^{n-2}a+\cdots+a^{n-1}\right) = \underbrace{a^{n-1}+a^{n-1}+\cdots+a^{n-1}}_{n \text{ terms}} = na^{n-1}

So $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$ , as required. The same result extends to rational $n$ (with $a>0$ when $n$ is not an integer), where the conjugate-multiplication technique plays the role that factorisation played here.

Worked example

Evaluate $\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}$ , then confirm the answer a second way using $\displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}$ .

Step 1 — try direct substitution on the first limit. At $x=3$ : numerator $=9-9=0$ , denominator $=3-3=0$ , giving the indeterminate form $\tfrac{0}{0}$ — substitution alone gives no answer here.

Step 2 — factorise and cancel. $x^2-9=(x-3)(x+3)$ , so $\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3$ , valid since $x \ne 3$ throughout the limiting process.

Step 3 — substitute in the simplified expression. $\lim_{x\to 3}(x+3) = 3+3 = 6$ . This matches the standard formula directly: here $n=2$ , $a=3$ , so $na^{n-1}=2\cdot3=6$ . $\lim_{x\to 3}\dfrac{x^2-9}{x-3} = 6$ .

Step 4 — now the surd case. At $x=4$ : $\sqrt{4}-2=0$ and $4-4=0$ , again $\tfrac{0}{0}$ . Rationalise by multiplying by the conjugate $\sqrt{x}+2$ : $\dfrac{\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2}=\dfrac{x-4}{(x-4)(\sqrt{x}+2)}=\dfrac{1}{\sqrt{x}+2}$ after cancelling $(x-4)$ .

Step 5 — substitute. $\lim_{x\to 4}\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{\sqrt4+2}=\dfrac14$ . Checked against the standard formula with $n=\tfrac12$ , $a=4$ : $na^{n-1}=\tfrac12\cdot4^{-1/2}=\tfrac12\cdot\tfrac12=\tfrac14$ . $\lim_{x\to 4}\dfrac{\sqrt{x}-2}{x-4} = \dfrac14$ .

Competitive stretch

Evaluate $\displaystyle\lim_{x\to 2}\frac{x^3-8}{x^2-4}$ without long division: both the top and bottom fit the standard-limit pattern separately. Write $\dfrac{x^3-2^3}{x-2}\cdot\dfrac{x-2}{x^2-2^2}$ , so the limit is $\dfrac{\lim_{x\to2}(x^3-2^3)/(x-2)}{\lim_{x\to2}(x^2-2^2)/(x-2)} = \dfrac{3\cdot2^2}{2\cdot2^1}=\dfrac{12}{4}=3$ — two applications of $na^{n-1}$ instead of one, avoiding a longer factorisation by inspection.

Common traps

Writing " $\tfrac{0}{0}=0$ " or " $\tfrac{0}{0}=1$ " after substitution — $\tfrac00$ is not a number; it is only a signal to factorise or rationalise.
Cancelling $(x-a)$ before actually confirming it is a genuine common factor of both numerator and denominator — an incomplete or wrong factorisation gives a wrong limit.
Forgetting why cancelling $(x-a)$ is allowed: it is because $x$ never equals $a$ during the approach ( $x-a \ne 0$ ), even though $a$ itself is the target point.
Forcing the formula $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$ onto an expression that does not literally match that shape, instead of first massaging the algebra (factoring out constants, splitting a compound expression) into that exact form.

Ask the tutor: ask your AI tutor for a fresh rational or surd limit and to check whether you should factorise, rationalise, or reach directly for the $na^{n-1}$ formula.

Why does cancelling x minus a not change the value of the limit, even though we are dividing by something that becomes zero at x equals a?

Because the limit never actually evaluates the expression at $x=a$ — it studies what happens as $x$ gets arbitrarily close to $a$ while staying different from it. For every such $x$ , $x - a$ is a genuine nonzero number, so dividing both numerator and denominator by it is perfectly ordinary algebra. Only after the cancelled, simplified expression is obtained do we substitute $x=a$ , and by then the problematic factor is already gone.

How do I know whether to factorise or to rationalise for a given 0 over 0 limit?

Look at what is causing the indeterminate form. If the numerator and denominator are both ordinary polynomial expressions, factorise them and cancel the shared linear factor. If a square root (or other surd) appears in the numerator or denominator, rationalise by multiplying top and bottom by the conjugate of the surd expression — this converts the surd into a plain polynomial difference, which can then be cancelled exactly as in the factorisation case.

Does the formula lim x to a of x to the n minus a to the n over x minus a require n to be a positive whole number?

Not in the version stated for this syllabus — $n$ may be any rational number, as long as $a>0$ when $n$ is not an integer (so that $a^n$ makes sense as a real number). For positive integer $n$ the result follows from ordinary polynomial factorisation, as derived above; for fractional $n$ , such as $n=\tfrac12$ giving a square root, the same value $na^{n-1}$ is reached by rationalising instead of factorising — both are the same underlying idea applied to different algebraic shapes.

Limits of algebraic functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads