MathsISC Class 11 · Limits and Derivatives

Limits: LHL and RHL, Algebra of Limits and the Standard Limits

Q: Why do we say x approaches a but need not equal a?

The limit describes the behaviour near a; the function may be undefined at a itself, as in (x² − 4)/(x − 2) at x = 2.

Q: Is infinity a limit?

A limit can be infinite in the sense of growing without bound, but then the limit does not exist as a real number.

Q: Can I use L'Hôpital's rule?

It is not part of this syllabus; use factorisation, rationalisation or standard limits.

Updated 17 October 2026 · Fact-checked

The limit of f(x) as x → a exists if the left-hand limit equals the right-hand limit. For indeterminate forms 0/0, factorise, rationalise or use standard limits such as (xⁿ − aⁿ)/(x − a) → naⁿ⁻¹, sin x / x → 1, (eˣ − 1)/x → 1 and log(1 + x)/x → 1 as x → 0.

Approach x = a from the left and the right; if both branches reach the same height L, the limit exists even though the point itself is a hole.

Illustration drawn for QuestionBuddy.

Understand Limits and Standard Limits

Idea of a limit. $lim_{x \to a} f (x) = L$ means f(x) gets as close as we please to L when x is close to a (but not equal to a). The left-hand limit (LHL) uses x < a; the right-hand limit (RHL) uses x > a. The limit exists only if LHL = RHL.

Algebra of limits. If the limits of f and g exist, then the limit of $f \pm g$ , $f g$ and $f / g$ (with the limit of g ≠ 0) is respectively the sum, difference, product and quotient of the limits.

Indeterminate forms. If substitution gives $0/0$ : (1) factorise and cancel; (2) rationalise (multiply by the conjugate); (3) use a standard limit.

Standard limits.

$lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$ .
$lim_{x \to 0} \frac{s i n x}{x} = 1$ , $lim_{x \to 0} \frac{t a n x}{x} = 1$ , $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2}$ (x in radians).
$lim_{x \to 0} \frac{e ^{x} - 1}{x} = 1$ , $lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a$ , $lim_{x \to 0} \frac{l n ( 1 + x )}{x} = 1$ .

Techniques. Reduce to a standard form by multiplying and dividing: $\frac{s i n 5 x}{3 x} = \frac{5}{3} \cdot \frac{s i n 5 x}{5 x}$ . Use $1 - cos 2 x = 2 sin^{2} x$ . Substitute $h = x - a$ to move the limit point to 0.

Presenting. Write the form obtained by substitution, state the method, and give the value, checking that the limit exists (LHL = RHL) where the function is piecewise or contains a modulus.

Limits and Standard Limits: formulas to remember

Power limit: $\lim_{x→a} \frac{x^n - a^n}{x - a} = n a^{n-1}$; n any rational number.
Trigonometric: $\lim_{x→0} \frac{\sin x}{x} = 1$ | $\lim_{x→0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$; x in radians.
Exponential and log: $\lim_{x→0} \frac{e^x - 1}{x} = 1$ | $\lim_{x→0} \frac{a^x - 1}{x} = \ln a$ | $\lim_{x→0} \frac{\ln(1+x)}{x} = 1$; Base e.
Existence: $L H L = R H L = L ⟹ l imi t i s L$; Check for piecewise and modulus functions.

Limits and Standard Limits in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Instantaneous speed from average speeds

Suppose a car's distance is $s = t^{2}$ metres at time t seconds, and we study the average speed between t = 2 and t = 2 + h.

Find the limit of the average speed as h → 0.

Try it first, then show the solutionSolution

Average speed $= \frac{( 2 + h ) ^{2} - 2 ^{2}}{h} = \frac{4 h + h ^{2}}{h} = 4 + h$ .
As h → 0 the average speed tends to 4.
So the instantaneous speed at t = 2 is 4 m s⁻¹.

Answer: 4 m s⁻¹.

Memory hook: Cancel the common factor h before letting h → 0.

Tricks to remember Limits and Standard Limits

Short memory aids. Say each one aloud once, then use it on a question the same day.

0/0 means work
Factorise or use a standard limit; never leave 0/0 as the answer.
Use it when: Indeterminate forms.
Match the argument
For sin(kx)/x, write it as k · sin(kx)/(kx).
Use it when: Trigonometric limits.

Limits and Standard Limits: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A trigonometric limit · 1 mark · MCQBoard level

The value of $lim_{x \to 0} \frac{s i n 3 x}{x}$ is:
(a) 1
(b) 3
(c) 1/3
(d) 0

Try it first, then show the solutionSolution

$\frac{s i n 3 x}{x} = 3 \cdot \frac{s i n 3 x}{3 x} \to 3$ .

Answer: (b) 3

Example 2: Limits by factorisation · 3 marks · Short answerBoard level

Evaluate $lim_{x \to 2} \frac{x ^{2} - 4}{x - 2}$ and $lim_{x \to 2} \frac{x ^{3} - 8}{x - 2}$ .

Try it first, then show the solutionSolution

$\frac{x ^{2} - 4}{x - 2} = x + 2 \to 4$ . (1 mark)
By the standard limit, $lim \frac{x ^{3} - 2 ^{3}}{x - 2} = 3 \cdot 2^{2} = 12$ . (1 mark)
Check: $x^{2} + 2 x + 4 \to 12$ at x = 2. (1 mark)

Answer: 4 and 12.

Example 3: Trigonometric limits · 4 marks · Short answerHarder board level

Evaluate $lim_{x \to 0} \frac{1 - c o s 2 x}{x ^{2}}$ and $lim_{x \to 0} \frac{s i n 5 x}{t a n 3 x}$ .

Try it first, then show the solutionSolution

$1 - cos 2 x = 2 sin^{2} x$ , so $\frac{2 s i n ^{2} x}{x ^{2}} \to 2$ . (2 marks)
$\frac{s i n 5 x}{t a n 3 x} = \frac{s i n 5 x}{5 x} \cdot \frac{3 x}{t a n 3 x} \cdot \frac{5}{3} \to \frac{5}{3}$ . (2 marks)

Answer: 2 and 5/3.

Example 4: Existence of a limit · 4 marks · Short answerHarder board level

Examine whether $lim_{x \to 0} \frac{∣ x ∣}{x}$ exists. For $f (x) = x^{2} + 1$ if $x < 1$ and $f (x) = 3 x - 1$ if $x \geq 1$ , find the limit at x = 1.

Try it first, then show the solutionSolution

For x > 0, $\frac{∣ x ∣}{x} = 1$ (RHL = 1); for x < 0, it is −1 (LHL = −1). (1.5 marks)
LHL ≠ RHL, so the limit does not exist. (0.5 mark)
At x = 1: LHL $= 1 + 1 = 2$ and RHL $= 3 - 1 = 2$ . (1.5 marks)
Both equal 2, so the limit is 2. (0.5 mark)

Answer: The first limit does not exist; the second equals 2.

Example 5: Mixed limits · 5 marks · Long answerHOTS / competency

Evaluate $lim_{x \to 0} \frac{e ^{3 x} - 1}{s i n 2 x}$ , $lim_{x \to 0} \frac{1 + x - 1}{x}$ and $lim_{x \to 0} \frac{2 ^{x} - 1}{x}$ .

Try it first, then show the solutionSolution

$\frac{e ^{3 x} - 1}{3 x} \cdot \frac{2 x}{s i n 2 x} \cdot \frac{3}{2} \to \frac{3}{2}$ . (1.5 marks)
Rationalise: $\frac{1 + x - 1}{x} \cdot \frac{1 + x + 1}{1 + x + 1} = \frac{1}{1 + x + 1} \to \frac{1}{2}$ . (2 marks)
$lim \frac{2 ^{x} - 1}{x} = ln 2 \approx 0.693$ . (1.5 marks)

Answer: 3/2; 1/2; ln 2.

Where students lose marks in Limits and Standard Limits

The slips examiners see most often, and what to write instead.

Mistake: Substituting without checking the form
Writing 0/0 = 0 or 1.
Do this: 0/0 is indeterminate: factorise or use a standard limit.
Usually costs: 2 marks.
Mistake: Degrees in trigonometric limits
Using sin x/x → 1 with x in degrees.
Do this: The standard limit needs radians.
Usually costs: 1 to 2 marks.
Mistake: Skipping LHL and RHL for modulus functions
Concluding without checking both sides.
Do this: The limit exists only if LHL = RHL.
Usually costs: 2 marks.

Exam tips

State the form first (0/0, etc.).
Match the argument inside sin, tan, e, ln with the denominator.
Rationalise for square roots.
Check LHL and RHL for piecewise functions.

Practice questions on Limits and Standard Limits

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $lim_{x \to 0} \frac{t a n x}{x}$ is:
(a) 0
(b) 1
(c) ∞
(d) does not exist
Hint
A standard limit.
Answer
(b) 1
Q2 · 2 marks · Short answerBoard level
Evaluate $lim_{x \to 3} \frac{x ^{2} - 9}{x - 3}$ .
Hint
Factorise.
Answer
$x + 3 \to 6$ .
Q3 · 3 marks · Short answerHarder board level
Evaluate $lim_{x \to 0} \frac{s i n 4 x}{s i n 2 x}$ .
Hint
Use 2 sin x cos x.
Answer
$\frac{s i n 4 x}{4 x} \cdot \frac{2 x}{s i n 2 x} \cdot 2 \to 2$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $lim_{x \to 1} \frac{x ^{4} - 1}{x - 1}$ and $lim_{x \to a} \frac{x ^{7} - a ^{7}}{x - a}$ .
Hint
Power-limit formula.
Answer
$4 \cdot 1^{3} = 4$ ; $7 a^{6}$ .
Q5 · 3 marks · Short answerHarder board level
Find $lim_{x \to 0} \frac{l n ( 1 + 4 x )}{x}$ and $lim_{x \to 0} \frac{e ^{2 x} - 1}{x}$ .
Hint
Match the arguments.
Answer
$4 \cdot \frac{l n ( 1 + 4 x )}{4 x} \to 4$ ; $2 \cdot \frac{e ^{2 x} - 1}{2 x} \to 2$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the average growth rate of a plant between t and t + h days is $\frac{( t + h ) ^{3} - t ^{3}}{h}$ cm per day.
(i) Simplify the expression. (1 mark)
(ii) Find the limit as h → 0. (1 mark)
(iii) Evaluate it at t = 2. (1 mark)
(iv) State what the limit represents. (1 mark)
Hint
Expand (t + h)³.
Answer
(i) $3 t^{2} + 3 t h + h^{2}$ . (ii) $3 t^{2}$ . (iii) $12$ cm per day. (iv) The instantaneous growth rate at time t.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.lim sin x/x?Show answer
1.
2.lim (eˣ − 1)/x?Show answer
1.
3.lim (xⁿ − aⁿ)/(x − a)?Show answer
n aⁿ⁻¹.
4.lim (1 − cos x)/x²?Show answer
1/2.

Limits and Standard Limits: frequently asked questions

Why do we say x approaches a but need not equal a?

The limit describes the behaviour near a; the function may be undefined at a itself, as in (x² − 4)/(x − 2) at x = 2.

Is infinity a limit?

A limit can be infinite in the sense of growing without bound, but then the limit does not exist as a real number.

Can I use L'Hôpital's rule?

It is not part of this syllabus; use factorisation, rationalisation or standard limits.