MHT-CET Maths · Limits

Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

When substitution gives 0/0 in an algebraic expression, a hidden factor of (x − a) is cancelling — factor it out, rationalise it out, or quote the xⁿ − aⁿ standard form.

All Limits notes MHT-CET Maths strategy

Why this matters

16 PYQs at 44% HARD, spread across every year from 2021 to 2026 — the most evenly recurring page in the chapter. Three stems here have been set twice in different sittings with the numbers unchanged, so the forms are worth knowing cold: a double rationalisation, a nested square root, and the 'limit is finite, find a and b' problem. The derivative-in-disguise reading at the end is the single fastest tool in the chapter and reappears on the continuity pages.

Concept 1 of 6: Spotting the 0/0 Form

Always substitute first. If you get a number, that is the answer. If you get

\dfrac{0}{0}

, the numerator and denominator share a factor that vanishes at

a

, and the whole game is to expose and cancel it.

Definition

Step 0 of every limit: substitute $x = a$ . A finite non-zero denominator means the limit is the value — stop.
$\dfrac{0}{0}$ is indeterminate: it tells you a factor $(x - a)$ is hiding on both floors, not what the answer is.
$\dfrac{\text{non-zero}}{0}$ is not indeterminate: the function blows up and there is no finite limit (check the sign of each side if the question asks).
The three algebraic tools for $0/0$ , in the order to try them: factor (polynomials), rationalise (square roots), quote the standard form $\dfrac{x^n - a^n}{x - a}$ (fractional or large powers).
If $x = a$ makes a polynomial zero, then $(x - a)$ divides it exactly — the factor theorem is what makes the cancellation possible.

Worked example

Evaluate

\lim_{x\to 3}\dfrac{x^2 - 9}{x - 3}

Practise this concept4 quick reps

Cancelling before checking the form

Cancelling

(x - a)

is only valid when it is genuinely a factor of both floors. If substitution gives

\dfrac{5}{0}

there is nothing to cancel and no finite limit — writing an option like '5' or '0' there is the standard distractor.

Concept 2 of 6: Factor and Cancel

Polynomials vanish at

a

because

(x - a)

divides them. Pull that factor out of top and bottom, cancel, and substitute into what is left.

Definition

Know the factorisations by heart: $x^2 - a^2 = (x - a)(x + a)$ , $x^3 - a^3 = (x - a)(x^2 + ax + a^2)$ , $x^3 + a^3 = (x + a)(x^2 - ax + a^2)$ .
For a general polynomial that vanishes at $a$ , divide by $(x - a)$ (synthetic division) to expose the factor.
Irrational points work identically: $x^4 - 4 = (x^2 - 2)(x^2 + 2) = (x - \sqrt2)(x + \sqrt2)(x^2 + 2)$ .
A difference of two fractions that each blow up must be combined into one fraction first; only the combined numerator has the cancelling factor.
Sometimes the point $a$ is itself hidden — given as 'where $f$ attains its maximum' or as a computed product — resolve it before touching the limit.

Factorisations that unlock 0/0

x^2 - a^2 = (x - a)(x + a) \qquad x^3 - a^3 = (x - a)(x^2 + ax + a^2) \qquad p(a) = 0 \Rightarrow (x - a) \mid p(x)

Worked example

Evaluate

\lim_{x\to 1}\dfrac{x^3 - 1}{x^2 - 1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q131Moderate

Example 2 · Limits · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

Evaluate

\lim_{x \to \sqrt{2}} \dfrac{x^{4} - 4}{x^{2} + 3\sqrt{2}\,x - 8}

Two blowing-up fractions must be combined first

\dfrac{1}{x - 2} - \dfrac{2x}{x^3 - 3x^2 + 2x}

x \to 2

: each piece alone is

\infty

, and '

\infty - \infty = 0

' is a trap. Factor the second denominator as

x(x - 1)(x - 2)

, put everything over it, and the combined numerator

(x - 2)(x + 1)

cancels to give

\dfrac{3}{2}

Drill 2 more on factor and cancel

Concept 3 of 6: Rationalisation — Single, Double and Nested Surds

A square root hides its factor of

(x - a)

. Multiplying by the conjugate

\sqrt{A} + \sqrt{B}

converts

\sqrt{A} - \sqrt{B}

into

A - B

, which is a polynomial that can be factored like any other.

Definition

Conjugate rule: $\sqrt{A} - \sqrt{B} = \dfrac{A - B}{\sqrt{A} + \sqrt{B}}$ . The new denominator is harmless — it does not vanish at $a$ .
Surds in both numerator and denominator: rationalise both, one after the other. Each conjugate contributes a factor to evaluate at the end.
Nested roots $\sqrt{1 + \sqrt{1 + u}} - \sqrt2$ : rationalise the outer difference to get $\sqrt{1 + u} - 1$ on top, then rationalise that.
After each rationalisation, substitute into the conjugate factors immediately — they are continuous at $a$ — and keep only the part that is still $0/0$ .

Conjugate rule

\frac{\sqrt{A} - \sqrt{B}}{C} = \frac{A - B}{C\left(\sqrt{A} + \sqrt{B}\right)}

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\sqrt{1 + x} - 1}{x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q124Hard

Example 3 · Limits · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

\displaystyle\lim_{x\to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}=

Rationalising only one floor when both carry surds

\dfrac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}

the top rationalises to

a - x

and the bottom to

3(a - x)

; stopping after one of them leaves a

0/0

you cannot substitute into. Do both, cancel

(a - x)

, then evaluate the two conjugate factors at

x = a

to get

\dfrac{2}{3\sqrt3}

Drill 2 more on rationalisation — single, double and nested surds

Concept 4 of 6: The xⁿ − aⁿ Standard Form and Fractional Powers

\dfrac{x^n - a^n}{x - a}

is the slope of

x^n

a

, and the result

na^{n-1}

holds for ANY rational

n

— which is what makes cube roots and fourth roots tractable without a conjugate.

Definition

$\lim_{x\to a}\dfrac{x^n - a^n}{x - a} = na^{n-1}$ for every rational $n$ , positive or negative, integer or fraction.
To use it, rewrite the root as a power: $(84 - x)^{1/4} - 3$ at $x \to 3$ is $u^{1/4} - 81^{1/4}$ with $u = 84 - x \to 81$ ; note $x - 3 = -(u - 81)$ , which supplies a minus sign.
Ratio of two such forms: $\dfrac{x^m - a^m}{x^n - a^n} = \dfrac{m}{n}a^{m-n}$ — divide numerator and denominator by $(x - a)$ .
Small- $x$ version: $\lim_{x\to 0}\dfrac{(1 + x)^n - 1}{x} = n$ , i.e. $(1 + x)^n \approx 1 + nx$ for small $x$ . This is how $\sqrt[3]{1 + a} - 1 \approx \dfrac{a}{3}$ , $\sqrt{1 + a} - 1 \approx \dfrac{a}{2}$ , $\sqrt[6]{1 + a} - 1 \approx \dfrac{a}{6}$ turn a frightening quadratic into $2x^2 + 3x + 1 = 0$ .

The xⁿ − aⁿ family

\lim_{x\to a}\frac{x^n - a^n}{x - a} = n a^{n-1} \quad (n \in \mathbb{Q}) \qquad \lim_{x\to 0}\frac{(1 + x)^n - 1}{x} = n

Worked example

Evaluate

\lim_{x\to 8}\dfrac{x^{1/3} - 2}{x - 8}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q108Moderate

Example 4 · Limits · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

\underset{x\rightarrow 3}{\lim} \frac{(84 -x)^{\frac{1}{4}}- 3}{x- 3}

The inner function's sign

\dfrac{(84 - x)^{1/4} - 3}{x - 3}

the inner variable is

84 - x

, which DECREASES as

x

increases. Writing

u = 84 - x

gives

x - 3 = -(u - 81)

, so the answer is

-\dfrac{1}{4}\cdot 81^{-3/4} = -\dfrac{1}{108}

, not

+\dfrac{1}{108}

. Both signs are always in the options.

Drill 1 more on the xⁿ − aⁿ standard form and fractional powers

Concept 5 of 6: The Derivative in Disguise — [f(x) − f(a)]/(x − a) and L'Hôpital

\dfrac{f(x) - f(a)}{x - a}

is the definition of

f'(a)

. Any

0/0

limit whose denominator is

x - a

and whose numerator vanishes at

a

can be read as a derivative, and answered by differentiating instead of factoring.

Definition

Definition of the derivative: $\lim_{x\to a}\dfrac{f(x) - f(a)}{x - a} = f'(a)$ . A polynomial numerator $p(x) - p(a)$ over $x - a$ is therefore $p'(a)$ — no factoring needed.
L'Hôpital's rule: if $\dfrac{f(x)}{g(x)}$ is $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ at $a$ , then $\lim\dfrac{f}{g} = \lim\dfrac{f'}{g'}$ whenever the right side exists. Differentiate top and bottom separately — this is not the quotient rule.
Apply it only to an indeterminate form; on $\dfrac{3}{5}$ it produces nonsense.
When the stem gives $f(a), f'(a), g(a), g'(a)$ as numbers and asks for a limit at $a$ , the question is L'Hôpital by construction: differentiate and substitute the given values.
A limit of the form $\dfrac{\int_{c}^{h(x)} \phi(t)\,dt}{x - a}$ is the same idea: the derivative of the integral is $\phi(h(x))\,h'(x)$ .

Derivative form and L'Hôpital

\lim_{x\to a}\frac{f(x) - f(a)}{x - a} = f'(a) \qquad \lim_{x\to a}\frac{f(x)}{g(x)} \overset{0/0}{=} \lim_{x\to a}\frac{f'(x)}{g'(x)}

Worked example

Evaluate

\lim_{x\to 2}\dfrac{x^5 + x^3 - 40}{x - 2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q117Moderate

Example 5 · Limits · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

f(a)=2, f'(a)=1, g(a)=-1, g'(a)=2

, then as

x

approaches

a

\frac{g(x)f(a) - g(a)f(x)}{x-a}

approaches

L'Hôpital on a form that is not indeterminate

\lim_{x\to 1}\dfrac{x^2 + 1}{x + 1}

1

by substitution; 'differentiating' gives

\dfrac{2x}{1} \to 2

, which is wrong. The rule has a precondition, and the paper's distractors are built from students who skip it.

Drill 1 more on the derivative in disguise — [f(x) − f(a)]/(x − a) and l'hôpital

Concept 6 of 6: A Finite Limit Forces the Numerator to Vanish — Finding a and b

\dfrac{x^2 - ax + b}{x - 1}

has a finite limit at

1

, the numerator must be

0

1

— otherwise the fraction blows up. That hidden equation is the one students forget, and without it there are two unknowns and one equation.

Definition

Condition 1: $\lim_{x\to a}\dfrac{p(x)}{x - a}$ is finite only if $p(a) = 0$ . Write this equation down first.
Condition 2: with $p(a) = 0$ the limit is $p'(a)$ (the derivative-in-disguise reading), which the stem sets equal to the given value.
Two equations, two unknowns — solve, and answer the combination asked for ( $a + b$ , $ab$ , and so on).
Same logic with a general denominator $q(x)$ that vanishes at $a$ : the numerator must share the factor.

Finite limit at a zero of the denominator

\lim_{x\to a}\frac{p(x)}{x - a} = L \text{ (finite)} \ \Rightarrow\ p(a) = 0 \ \text{ and } \ p'(a) = L

Worked example

\lim_{x\to 2}\dfrac{x^2 + ax + b}{x - 2} = 3

, find

a + b

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q134Moderate

Example 6 · Limits · Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

\lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 5

, then

(a+b)

is equal to

Treating a as free and reading b off the limit

Skipping

p(a) = 0

leaves one equation for two unknowns, and every option looks reachable. The vanishing condition is not optional — it is the reason the limit is finite at all.

Drill 1 more on a finite limit forces the numerator to vanish — finding a and b

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Factor and Cancel
Factorisations that unlock 0/0
$x^2 - a^2 = (x - a)(x + a) \qquad x^3 - a^3 = (x - a)(x^2 + ax + a^2) \qquad p(a) = 0 \Rightarrow (x - a) \mid p(x)$
Rationalisation — Single, Double and Nested Surds
Conjugate rule
$\frac{\sqrt{A} - \sqrt{B}}{C} = \frac{A - B}{C\left(\sqrt{A} + \sqrt{B}\right)}$
The xⁿ − aⁿ Standard Form and Fractional Powers
The xⁿ − aⁿ family
$\lim_{x\to a}\frac{x^n - a^n}{x - a} = n a^{n-1} \quad (n \in \mathbb{Q}) \qquad \lim_{x\to 0}\frac{(1 + x)^n - 1}{x} = n$
The Derivative in Disguise — [f(x) − f(a)]/(x − a) and L'Hôpital
Derivative form and L'Hôpital
$\lim_{x\to a}\frac{f(x) - f(a)}{x - a} = f'(a) \qquad \lim_{x\to a}\frac{f(x)}{g(x)} \overset{0/0}{=} \lim_{x\to a}\frac{f'(x)}{g'(x)}$
A Finite Limit Forces the Numerator to Vanish — Finding a and b
Finite limit at a zero of the denominator
$\lim_{x\to a}\frac{p(x)}{x - a} = L \text{ (finite)} \ \Rightarrow\ p(a) = 0 \ \text{ and } \ p'(a) = L$

Watch out for (6)

Cancelling before checking the form
→ Spotting the 0/0 Form
Two blowing-up fractions must be combined first
→ Factor and Cancel
Rationalising only one floor when both carry surds
→ Rationalisation — Single, Double and Nested Surds
The inner function's sign
→ The xⁿ − aⁿ Standard Form and Fractional Powers
L'Hôpital on a form that is not indeterminate
→ The Derivative in Disguise — [f(x) − f(a)]/(x − a) and L'Hôpital
Treating a as free and reading b off the limit
→ A Finite Limit Forces the Numerator to Vanish — Finding a and b

Test yourself on Limits

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Limits chapter test Drill all 16 questions

Related notes

Differentiation — Foundations, Chain Rule & Differentiability (the derivative this page keeps borrowing) →