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

13 PYQs at 46% HARD, spread across every year from 2021 to 2025 — 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

Intuition

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}

Practice 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

Intuition

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}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2LimitsMODERATE

Evaluate

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

[Q131 · May Shift 1 · 2021]

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

Intuition

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}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3LimitsHARD

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

[Q124 · 10th May Shift 2 · 2024]

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 3 more on rationalisation — single, double and nested surds

Concept 4 of 6

The xⁿ − aⁿ Standard Form and Fractional Powers

Intuition

\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}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4LimitsMODERATE

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

[Q108 · 19 April Shift I · 2025]

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

Intuition

\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}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5LimitsMODERATE

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

[Q117 · 14th May Shift 2 · 2024]

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

Intuition

\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

Practice this conceptself-check

From the bank · past-year question

Example 6LimitsMODERATE

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

, then

(a+b)

is equal to

[Q134 · Shift 1 · 2022]

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

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Related notes

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