NDA Maths · Limits & Continuity

Evaluating Limits — Factor, Rationalise, Standard Forms

The core toolkit for a 0/0 limit: direct substitution first, then factor-and-cancel, rationalise a surd, apply a standard limit, use L'Hôpital, or handle the 1^∞ form via the exponential trick.

All Limits & Continuity notes NDA Maths strategy

Why this matters

Most limit questions are a single recognition: which tool clears the indeterminate form. Get the standard limits cold and the 0/0 questions become one or two lines.

Concept 1 of 5

What a limit is — and when it exists

Intuition

A limit describes the value a function approaches, not necessarily its value at the point. It exists only when both sides agree — the left-hand limit equals the right-hand limit. Always try direct substitution first; only if that gives an indeterminate form (0/0, ∞/∞, 1^∞, …) do you need a technique.

Definition

$\lim_{x\to a}f(x)=L$ means $f(x)$ gets arbitrarily close to $L$ as $x\to a$ . It exists iff LHL $=$ RHL. Algebra of limits: limits distribute over sums, products, and quotients (when denominators are non-zero). For $n\to\infty$ ratios, the dominant term decides (e.g. $\dfrac{a^n+b^n}{a^{n}}\to$ the larger base's contribution).

Worked example

Evaluate

\lim_{x\to 2}(x^2+3x-1)

Practice this conceptself-check · 4 quick reps

Drill 4 more on what a limit is — and when it exists

Concept 2 of 5

0/0 by factoring, cancelling, and rationalising

Intuition

When substitution gives 0/0, the zero factor is shared by numerator and denominator. Factor and cancel it (often via

x^n-a^n

), or — when surds are involved — rationalise by multiplying by the conjugate to expose the cancelling factor.

Definition

Factor/cancel: use $x^n-a^n=(x-a)(x^{n-1}+\cdots+a^{n-1})$ ; the standard result $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$ .
Rationalise: multiply numerator and denominator by the conjugate of the surd to turn $\sqrt{A}-\sqrt{B}$ into $A-B$ , then cancel.

The x^n − a^n standard limit

\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}

Worked example

Evaluate

\lim_{x\to 2}\dfrac{x^2-4}{x-2}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Limits & ContinuityEASY

What is

\lim_{x\to1}\dfrac{x^{9}-1}{x^{3}-1}

equal to?

[Q100 · Apr · 2023]

Drill 12 more on 0/0 by factoring, cancelling, and rationalising

Concept 3 of 5

The standard limits to memorise

Intuition

A fixed list of limits resolves most trigonometric, exponential, and logarithmic 0/0 forms instantly. Reduce the problem to one of these (often by multiplying/dividing by the right variable) and read off the value.

Definition

Each holds as $x\to 0$ (angles in radians). For a scaled argument, the factor scales too — e.g. $\lim_{x\to0}\dfrac{\sin ax}{x}=a$ .

Limit (as x → 0)	Value
sin x / x	1
tan x / x	1
(1 − cos x) / x²	1/2
log(1 + x) / x	1
(eˣ − 1) / x	1
(aˣ − 1) / x	ln a
(1 + x)^(1/x)	e

Radians only. Scale the argument and the value scales: sin(ax)/x → a.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Limits & ContinuityEASY

What is

\lim_{x\to0}\dfrac{10^{\sin x}-1}{\tan x}

equal to?

[Q73 · Apr · 2026]

sin x / x → 1 only as x → 0

The standard limit

\dfrac{\sin x}{x}\to 1

holds only as

x\to 0

. As

x\to\pi

\dfrac{\sin x}{x}

is just

\dfrac{\sin\pi}{\pi}=0

by direct substitution — there is no indeterminate form, so the standard result does not apply.

Scale the argument, scale the value

\dfrac{\sin ax}{x}\to a

, not

1

— and

\dfrac{\sin ax}{bx}\to\dfrac{a}{b}

. Forgetting the multiplier is the most common slip:

\dfrac{\sin 5x}{x}\to 5

, and

\dfrac{\tan 3x}{\sin 2x}\to\dfrac{3}{2}

Drill 8 more on the standard limits to memorise

Concept 4 of 5

L'Hôpital's rule for 0/0 and ∞/∞

Intuition

When a limit is genuinely 0/0 or ∞/∞ and factoring is awkward, differentiate the top and bottom separately and try the limit again. Repeat if it's still indeterminate.

Definition

If $\lim\dfrac{f}{g}$ is $\tfrac00$ or $\tfrac{\infty}{\infty}$ and $f,g$ are differentiable, then $\lim\dfrac{f}{g}=\lim\dfrac{f'}{g'}$ (provided the latter exists). Only apply it to a true indeterminate form — never to a determinate one. Series expansion ( $e^x=1+x+\tfrac{x^2}{2}+\cdots$ , $\sin x=x-\tfrac{x^3}{6}+\cdots$ ) often does the same job faster.

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\sin x - x}{x^3}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Limits & ContinuityEASY

What is

\lim_{x\to 0}\dfrac{e^x-(1+x)}{x^2}

equal to?

[Q71 · Apr · 2017]

L'Hôpital only on indeterminate forms

Differentiating top and bottom is valid only when the limit is genuinely

\tfrac00

\tfrac{\infty}{\infty}

. Applying it to a determinate form like

\dfrac{2}{3}

\dfrac{\sin x}{x+1}

x=0

(which is

\tfrac01=0

) gives a WRONG answer.

Differentiate top and bottom separately — not the quotient

L'Hôpital replaces

\dfrac{f}{g}

\dfrac{f'}{g'}

— you do NOT apply the quotient rule. Differentiating

\dfrac{f}{g}

as a single function via the quotient rule is a different (and wrong) operation here.

Drill 2 more on l'hôpital's rule for 0/0 and ∞/∞

Concept 5 of 5

The 1^∞ form

Intuition

A limit of the shape

[f(x)]^{g(x)}

where

f\to 1

and

g\to\infty

is the indeterminate

1^\infty

. Take logs (or use the standard shortcut) to convert it to a 0·∞ product, which becomes an ordinary limit in the exponent.

Definition

If $f(x)\to 1$ and $g(x)\to\infty$ , then $\lim [f(x)]^{g(x)}=e^{\,\lim\, g(x)\,[f(x)-1]}$ . Equivalently, $L=e^{\lim g\ln f}$ — take $\ln$ , evaluate the resulting product, then exponentiate.

The 1^∞ shortcut

\lim [f(x)]^{g(x)} = e^{\,\lim\, g(x)\,[f(x)-1]}\quad (f\to 1,\ g\to\infty)

Worked example

Evaluate

\lim_{x\to 0}(1+2x)^{1/x}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Limits & ContinuityHARD

Let

f(x)=\dfrac{a^{x-1}+b^{x-1}}{2}

and

g(x)=x-1

What is

\lim_{x\to1}f(x)^{1/g(x)}

equal to?

[Q54 · Sep · 2023]

Drill 1 more on the 1^∞ form

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 (2)

0/0 by factoring, cancelling, and rationalising
The x^n − a^n standard limit
$\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$
The 1^∞ form
The 1^∞ shortcut
$\lim [f(x)]^{g(x)} = e^{\,\lim\, g(x)\,[f(x)-1]}\quad (f\to 1,\ g\to\infty)$

Reference tables (1)

The standard limits to memorise7 rows

Limit (as x → 0)	Value
sin x / x	1
tan x / x	1
(1 − cos x) / x²	1/2
log(1 + x) / x	1
(eˣ − 1) / x	1
(aˣ − 1) / x	ln a
(1 + x)^(1/x)	e

Radians only. Scale the argument and the value scales: sin(ax)/x → a.

Watch out for (4)

sin x / x → 1 only as x → 0
→ The standard limits to memorise
Scale the argument, scale the value
→ The standard limits to memorise
L'Hôpital only on indeterminate forms
→ L'Hôpital's rule for 0/0 and ∞/∞
Differentiate top and bottom separately — not the quotient
→ L'Hôpital's rule for 0/0 and ∞/∞

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