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.

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

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

Worked example

Evaluate limx2(x2+3x1)\lim_{x\to 2}(x^2+3x-1).
  1. The function is a polynomial (continuous), so substitute directly.
  2. 22+3(2)1=4+61=92^2+3(2)-1=4+6-1=9.
Answer:99.
Practice this conceptself-check · 4 quick reps

Try it yourself

Does limx0xx\lim_{x\to 0}\dfrac{|x|}{x} exist?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    A limit exists iff?
  2. 2.
    First thing to try on any limit?
  3. 3.
    Is limxaf(x)\lim_{x\to a}f(x) necessarily f(a)f(a)?
  4. 4.
    limx1(x2+1)\lim_{x\to 1}(x^2+1)?

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 xnanx^n-a^n), or — when surds are involved — rationalise by multiplying by the conjugate to expose the cancelling factor.

Definition

  • Factor/cancel: use xnan=(xa)(xn1++an1)x^n-a^n=(x-a)(x^{n-1}+\cdots+a^{n-1}); the standard result limxaxnanxa=nan1\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 AB\sqrt{A}-\sqrt{B} into ABA-B, then cancel.

Worked example

Evaluate limx2x24x2\lim_{x\to 2}\dfrac{x^2-4}{x-2}.
  1. Factor: (x2)(x+2)x2=x+2\dfrac{(x-2)(x+2)}{x-2}=x+2 for x2x\neq 2.
  2. Substitute: 2+2=42+2=4.
Answer:44.
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate limx1x91x31\lim_{x\to 1}\dfrac{x^9-1}{x^3-1}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    limxaxnanxa=?\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=?
  2. 2.
    Tool for xaxa\dfrac{\sqrt{x}-\sqrt{a}}{x-a}?
  3. 3.
    limx1x41x1\lim_{x\to1}\dfrac{x^4-1}{x-1}?
  4. 4.
    limx1x91x31\lim_{x\to1}\dfrac{x^9-1}{x^3-1}?

From the bank · past-year question

Example 2Limits & ContinuityEASY
What is limx1x91x31\lim_{x\to1}\dfrac{x^{9}-1}{x^{3}-1} equal to?

[Q100 · Apr · 2023]

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 x0x\to 0 (angles in radians). For a scaled argument, the factor scales too — e.g. limx0sinaxx=a\lim_{x\to0}\dfrac{\sin ax}{x}=a.

Limit (as x → 0)Value
sin x / x1
tan x / x1
(1 − cos x) / x²1/2
log(1 + x) / x1
(eˣ − 1) / x1
(aˣ − 1) / xln 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

Try it yourself

Evaluate limx010sinx1tanx\lim_{x\to 0}\dfrac{10^{\sin x}-1}{\tan x}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    limx0sinxx\lim_{x\to0}\dfrac{\sin x}{x}?
  2. 2.
    limx01cosxx2\lim_{x\to0}\dfrac{1-\cos x}{x^2}?
  3. 3.
    limx0ax1x\lim_{x\to0}\dfrac{a^x-1}{x}?
  4. 4.
    limx0sin5xx\lim_{x\to0}\dfrac{\sin 5x}{x}?

From the bank · past-year question

Example 3Limits & ContinuityEASY
What is limx010sinx1tanx\lim_{x\to0}\dfrac{10^{\sin x}-1}{\tan x} equal to?

[Q73 · Apr · 2026]

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 limfg\lim\dfrac{f}{g} is 00\tfrac00 or \tfrac{\infty}{\infty} and f,gf,g are differentiable, then limfg=limfg\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 (ex=1+x+x22+e^x=1+x+\tfrac{x^2}{2}+\cdots, sinx=xx36+\sin x=x-\tfrac{x^3}{6}+\cdots) often does the same job faster.

Worked example

Evaluate limx0sinxxx3\lim_{x\to 0}\dfrac{\sin x - x}{x^3}.
  1. It is 0/0. Differentiate top and bottom: cosx13x2\dfrac{\cos x-1}{3x^2} (still 0/0).
  2. Again: sinx6x16\dfrac{-\sin x}{6x}\to-\tfrac16 (using sinx/x1\sin x/x\to 1).
Answer:16-\tfrac16.
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate limx0ex(1+x)x2\lim_{x\to 0}\dfrac{e^x-(1+x)}{x^2}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    L'Hôpital applies to which forms?
  2. 2.
    After differentiating top & bottom, then?
  3. 3.
    First-order expansion of exe^x?
  4. 4.
    Can you use it on 23\tfrac{2}{3}?

From the bank · past-year question

Example 4Limits & ContinuityEASY
What is limx0ex(1+x)x2\lim_{x\to 0}\dfrac{e^x-(1+x)}{x^2} equal to?

[Q71 · Apr · 2017]

Concept 5 of 5

The 1^∞ form

Intuition

A limit of the shape [f(x)]g(x)[f(x)]^{g(x)} where f1f\to 1 and gg\to\infty is the indeterminate 11^\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)1f(x)\to 1 and g(x)g(x)\to\infty, then lim[f(x)]g(x)=elimg(x)[f(x)1]\lim [f(x)]^{g(x)}=e^{\,\lim\, g(x)\,[f(x)-1]}. Equivalently, L=elimglnfL=e^{\lim g\ln f} — take ln\ln, evaluate the resulting product, then exponentiate.

The 1^∞ shortcut

lim[f(x)]g(x)=elimg(x)[f(x)1](f1, g)\lim [f(x)]^{g(x)} = e^{\,\lim\, g(x)\,[f(x)-1]}\quad (f\to 1,\ g\to\infty)

Worked example

Evaluate limx0(1+2x)1/x\lim_{x\to 0}(1+2x)^{1/x}.
  1. Form 11^\infty: use L=elimg[f1]L=e^{\lim g[f-1]} with f1=2xf-1=2x, g=1xg=\tfrac1x.
  2. Exponent =limx01x2x=2=\lim_{x\to0}\tfrac{1}{x}\cdot 2x=2.
Answer:e2e^2.
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate limx(1+3x)x\lim_{x\to\infty}\left(1+\tfrac{3}{x}\right)^{x}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    lim[f]g\lim[f]^{g} with f1,gf\to1,g\to\infty equals?
  2. 2.
    limx0(1+x)1/x\lim_{x\to0}(1+x)^{1/x}?
  3. 3.
    limx(1+2x)x\lim_{x\to\infty}(1+\tfrac2x)^{x}?
  4. 4.
    What indeterminate form is this?

From the bank · past-year question

Example 5Limits & ContinuityHARD
Let f(x)=ax1+bx1f(x)=a^{x-1}+b^{x-1} and g(x)=x1g(x)=\sqrt{x-1}.
What is limx1f(x)1/g(x)\lim_{x\to1}f(x)^{1/g(x)} equal to?

[Q54 · Sep · 2023]

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 (1)

  • The 1^∞ form

    The 1^∞ shortcut

    lim[f(x)]g(x)=elimg(x)[f(x)1](f1, g)\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 / x1
tan x / x1
(1 − cos x) / x²1/2
log(1 + x) / x1
(eˣ − 1) / x1
(aˣ − 1) / xln a
(1 + x)^(1/x)e
Radians only. Scale the argument and the value scales: sin(ax)/x → a.

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Limits & ContinuityEASY
Consider the following statements: 1. If limxaf(x)\lim_{x\to a}f(x) and limxag(x)\lim_{x\to a}g(x) both exist, then limxa{f(x)g(x)}\lim_{x\to a}\{f(x)g(x)\} exists. 2. If limxa{f(x)g(x)}\lim_{x\to a}\{f(x)g(x)\} exists, then both limxaf(x)\lim_{x\to a}f(x) and limxag(x)\lim_{x\to a}g(x) must exist. Which of the above statements is/are correct?

[Q93 · Apr · 2017]

Example 2Limits & ContinuityEASY
What is limx1x3+x2x2+3x+2\lim_{x\to -1}\frac{x^3+x^2}{x^2+3x+2} equal to?

[Q79 · Apr · 2021]

Example 3Limits & ContinuityEASY
What is the value of limx0sinxtan3x\lim_{x \to 0} \dfrac{\sin x^{\circ}}{\tan 3x^{\circ}} ?

[Q73 · Sep · 2019]

Example 4Limits & ContinuityHARD
If limxaaxxaxaaa=1\lim_{x\to a}\frac{a^x-x^a}{x^a-a^a}=-1, then what is the value of aa?

[Q76 · Apr · 2021]

Example 5Limits & ContinuityHARD
Let f(x)=ax1+bx1f(x)=a^{x-1}+b^{x-1} and g(x)=x1g(x)=\sqrt{x-1}.
What is limx1f(x)1g(x)\lim_{x\to1}\frac{f(x)-1}{g(x)} equal to?

[Q53 · Sep · 2023]

Drill every past-year question on this subtopic

31 questions from the bank — paginated, with cart and Word-export support.

Related notes