Quantitative Aptitude · Sets, Relations and Functions, Limits and Continuity

Limits and Standard Limit Formulas for CA Foundation

Updated 1 October 2026 · Fact-checked

A limit is the value a function approaches as x gets close to a point, not necessarily the value at that point. To solve it, substitute first. If you get 0/0, factorise, rationalise or use a standard formula. A limit exists only if left and right limits are equal.

Understand Limits and Standard Limit Formulas

A limit tells you what value f(x) heads towards as x comes very close to a number a. It does not ask what happens at x = a. It asks what happens near a.

Take f(x) = (x² − 1) ÷ (x − 1). At x = 1 it gives 0/0, so f(1) is undefined. But for x ≠ 1 the function equals x + 1. As x gets close to 1, f(x) gets close to 2. So the limit is 2, even though f(1) does not exist.

The left hand limit (LHL) is the value f(x) approaches as x comes towards a from values smaller than a. The right hand limit (RHL) is the value as x comes from values larger than a. The limit exists only when LHL = RHL, and both are finite. If they differ, the limit does not exist.

Most exam questions give 0/0 when you substitute. This is an indeterminate form. It does not mean the answer is zero or undefined. It means you must simplify first, by factorisation, rationalisation or a standard limit formula.

The standard limits involving eˣ, log and powers cover the cases where simple algebra will not work. Learn them as patterns. Then reshape the question until it matches one.

Key formulas to remember

Existence of a limit: lim(x→a) f(x) exists ⇔ LHL = RHL (both finite); LHL is the limit as x → a⁻. RHL is the limit as x → a⁺.
Sum and difference: lim [f(x) ± g(x)] = lim f(x) ± lim g(x); Valid when both individual limits exist.
Product and constant multiple: lim [f(x)·g(x)] = lim f(x) · lim g(x); lim [k·f(x)] = k · lim f(x); Valid when the individual limits exist.
Quotient: lim [f(x) ÷ g(x)] = lim f(x) ÷ lim g(x); Valid only when lim g(x) ≠ 0.
Power-difference limit: lim(x→a) (xⁿ − aⁿ) ÷ (x − a) = n·aⁿ⁻¹; Holds for any real n when a > 0. For a positive integer n it holds for all real a.
Exponential limit: lim(x→0) (eˣ − 1) ÷ x = 1; More generally, lim(x→0) (eᵃˣ − 1) ÷ x = a.
Logarithmic limit: lim(x→0) log(1 + x) ÷ x = 1; Here log means natural log (base e). More generally, lim(x→0) log(1 + ax) ÷ x = a.
General exponential limit: lim(x→0) (aˣ − 1) ÷ x = log a; Valid for a > 0. Log is natural log.
Definition of e: lim(x→0) (1 + x)^(1/x) = e; lim(x→∞) (1 + 1/x)ˣ = e; Useful for forms like (1 + kx)^(1/x), which tends to eᵏ as x → 0.
Limit at infinity of a ratio of polynomials: Compare the highest powers of x in numerator and denominator; Equal degree gives the ratio of leading coefficients. Higher degree in the denominator gives 0.

How to solve Limits and Standard Limit Formulas questions

Use this order for any limit question. Stop as soon as you get a finite answer.

1Substitute x = a directly. If you get a finite number, that is the answer.
2If you get 0/0, look at the type of expression. Polynomials point to factorisation. Square roots point to rationalisation. eˣ or log point to a standard formula.
3For factorisation, factorise numerator and denominator, cancel the common factor (x − a), then substitute again.
4For rationalisation, multiply the numerator and denominator by the conjugate of the root expression. Simplify, cancel (x − a), then substitute.
5For eˣ, aˣ or log forms, rewrite the expression so that the numerator and denominator match a standard pattern. Adjust by multiplying and dividing by the needed constant.
6For x → ∞, divide every term by the highest power of x in the denominator. Terms like 1/x tend to 0.
7For a piecewise function or modulus, find the LHL and RHL separately. If they are equal the limit exists. Otherwise it does not.
8Check your answer against the options. Confirm it is finite and matches the form you expected.

Quickest way: Substitute, spot the pattern, pick the option

When to use it: Use this in the MCQ paper when a limit gives 0/0 and you want the answer in under a minute.

Substitute the value at once. About half the questions end here.
For (xⁿ − aⁿ)/(x − a) types, write n·aⁿ⁻¹ directly. Do not factorise.
L'Hôpital's rule (ratio of derivatives) can be used to check a 0/0 form, but the syllabus method is factorisation or standard limits. Show the syllabus method in your working.
For eˣ and log questions, match the pattern and read off the constant. For example, (e³ˣ − 1)/x → 3, and log(1 + 5x)/x → 5.
For x → ∞, compare the highest powers only and ignore the rest.
Test options by plugging in a value close to a, such as a + 0.01, if you are stuck. Use this to eliminate wrong options, not to skip the method.
If a question needs more than two minutes, mark it and move on. A wrong answer costs 0.25 marks.

Common mistakes in Limits and Standard Limit Formulas

Writing the answer as 0 or 'undefined' when substitution gives 0/0.
Students treat 0/0 as a final result instead of a signal to simplify.
Fix: Treat 0/0 as a prompt. Factorise, rationalise or use a standard limit before substituting again.
Cancelling terms instead of factors, such as cancelling x in (x + 2)/(x + 3).
Rushing and treating addition like multiplication.
Fix: Factorise fully. Cancel only a factor common to the whole numerator and the whole denominator.
Using lim (eˣ − 1)/x = 1 when the exponent is 2x, and still answering 1.
Students memorise the base pattern and forget the coefficient adjustment.
Fix: Divide by the same expression as the exponent. (e²ˣ − 1)/x = 2 · (e²ˣ − 1)/(2x) → 2.
Saying a limit exists after checking only one side.
Students calculate the RHL, or substitute a value from one side, and stop.
Fix: Find both LHL and RHL. The limit exists only when they are equal and finite.
Confusing the limit with the function value, such as saying the limit is undefined because f(a) is undefined.
The limit is about nearby values, but students focus on the value at the point.
Fix: Remember f(a) can be undefined or different while the limit still exists.
Rationalising with the wrong conjugate, or forgetting to multiply the other part of the fraction.
Students multiply only the numerator or use the same sign instead of the opposite sign.
Fix: Flip the sign between the two terms. Multiply numerator and denominator by the same conjugate.

Worked examples

Example 1

The value of lim(x→2) (x² − 4) ÷ (x² − 3x + 2) is: (a) 0 (b) 2 (c) 4 (d) 1

Show the solution

Substitute x = 2. Numerator = 4 − 4 = 0. Denominator = 4 − 6 + 2 = 0. This is 0/0.
Factorise the numerator: x² − 4 = (x − 2)(x + 2).
Factorise the denominator: x² − 3x + 2 = (x − 1)(x − 2).
Cancel (x − 2). The expression becomes (x + 2) ÷ (x − 1).
Substitute x = 2: (2 + 2) ÷ (2 − 1) = 4 ÷ 1 = 4.

Answer: (c) 4

Example 2

The value of lim(x→0) (√(1 + x) − 1) ÷ x is: (a) 0 (b) 1/2 (c) 1 (d) 2

Show the solution

Substitute x = 0. Numerator = √1 − 1 = 0. Denominator = 0. This is 0/0.
Multiply numerator and denominator by the conjugate (√(1 + x) + 1).
Numerator becomes (1 + x) − 1 = x.
The expression is x ÷ [x(√(1 + x) + 1)]. Cancel x to get 1 ÷ (√(1 + x) + 1).
Substitute x = 0: 1 ÷ (1 + 1) = 1/2.

Answer: (b) 1/2

Example 3

The value of lim(x→0) (e⁵ˣ − 1) ÷ (e²ˣ − 1) is: (a) 1 (b) 2/5 (c) 5/2 (d) 0

Show the solution

Substitute x = 0. Numerator = 0 and denominator = 0. This is 0/0.
Divide the numerator and denominator by x. The expression becomes [(e⁵ˣ − 1) ÷ x] ÷ [(e²ˣ − 1) ÷ x].
Use lim (eᵃˣ − 1) ÷ x = a. The numerator tends to 5 and the denominator tends to 2.
The limit is 5 ÷ 2 = 5/2.

Answer: (c) 5/2

Exam tips

Always substitute first. Many MCQs are direct substitution questions placed among harder-looking ones.
Memorise the standard limits as patterns with a coefficient. Questions usually change the constant, such as e³ˣ or log(1 + 4x).
For left and right limit questions, check which side each piece of the function applies to before computing.
Keep conjugate pairs ready for questions with square roots. Practise the algebra until it takes under a minute.
Eliminate options that are impossible, such as an infinite value when the expression clearly simplifies to a finite one.

Practice questions from Sets, Relations and Functions, Limits and Continuity

Limits and Standard Limit Formulas: frequently asked questions

What is the difference between left hand limit and right hand limit?

The left hand limit is the value f(x) approaches as x nears a from below a. The right hand limit is the value as x nears a from above a. The limit exists only if both are equal.

How do I solve limits by factorisation and rationalisation?

Use factorisation when the expression is a ratio of polynomials giving 0/0. Cancel the common factor and substitute again. Use rationalisation when square roots are present. Multiply by the conjugate, simplify, cancel and substitute.

Which standard limit formulas should I learn for CA Foundation?

Learn (xⁿ − aⁿ)/(x − a) → n·aⁿ⁻¹, (eˣ − 1)/x → 1, log(1 + x)/x → 1, (aˣ − 1)/x → log a, and (1 + x)^(1/x) → e. Also learn the coefficient versions such as (eᵃˣ − 1)/x → a.

Can a limit exist if the function is not defined at that point?

Yes. A limit depends only on values near the point, not at it. For example, (x² − 1)/(x − 1) is undefined at x = 1 but its limit there is 2.