Quantitative Aptitude · Differential and Integral Calculus

Limits and Continuity Basics 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 limit. A function is continuous at a point if its limit there equals its value.

Understand Limits and Continuity Basics

A limit tells you where f(x) is heading as x gets very close to a number a. We write it as lim (x→a) f(x). The function does not need to be defined at a. We only care about values near a.

Take f(x) = (x² − 4) ÷ (x − 2). At x = 2 it gives 0/0, so f(2) is undefined. For x ≠ 2, it simplifies to x + 2. As x nears 2, f(x) nears 4. So the limit is 4, even though f(2) does not exist.

A limit exists only if the left-hand limit (x approaches a from below) equals the right-hand limit (x approaches a from above). If they differ, the limit does not exist.

A function is continuous at x = a if three things hold: f(a) is defined, the limit as x→a exists, and the limit equals f(a). In plain words, you can draw the graph through that point without lifting your pen.

Limit and continuity differ in one way. A limit looks at the neighbourhood of a. Continuity also demands that the actual value at a matches it. Every continuous point has a limit, but a point with a limit need not be continuous.

Key formulas to remember

Definition of limit: lim (x→a) f(x) = L, if LHL = RHL = L; LHL is the limit as x→a⁻ and RHL is the limit as x→a⁺. Both must be finite and equal.
Continuity at a point: lim (x→a) f(x) = f(a); f(a) must be defined and the limit must exist.
Limit laws: lim [f(x) ± g(x)] = lim f(x) ± lim g(x); lim [f(x) × g(x)] = lim f(x) × lim g(x); lim [f(x) ÷ g(x)] = lim f(x) ÷ lim g(x); Applies when each individual limit exists. For division, the limit of g(x) must not be 0.
Power standard limit: lim (x→a) (xⁿ − aⁿ) ÷ (x − a) = n·aⁿ⁻¹; Valid for any real n, with a > 0 when n is not an integer.
Exponential standard limit: lim (x→0) (eˣ − 1) ÷ x = 1; More generally, lim (x→0) (aˣ − 1) ÷ x = log_e a, for a > 0.
Logarithmic standard limit: lim (x→0) log_e(1 + x) ÷ x = 1; Natural log only.
Definition of e: lim (x→0) (1 + x)^(1/x) = e; Equivalent form: lim (n→∞) (1 + 1/n)ⁿ = e.
Limit at infinity: lim (x→∞) 1/xⁿ = 0, for n > 0; For a ratio of polynomials, divide the numerator and denominator by the highest power of x.

How to solve Limits and Continuity Basics questions

Use this order for any limit question. It stops you from wasting time on harder methods when substitution works.

1Substitute x = a directly into the expression.
2If you get a finite number, that is the limit. If the denominator is 0 and the numerator is not 0, the limit is not finite.
3If you get 0/0, factorise the numerator and denominator and cancel the common factor (x − a).
4If roots are present, multiply the numerator and denominator by the conjugate, then simplify and cancel.
5If the expression matches a standard form such as (xⁿ − aⁿ)/(x − a), (eˣ − 1)/x or log(1 + x)/x, apply the standard limit.
6If x→∞, divide every term by the highest power of x and use 1/xⁿ → 0.
7Substitute again into the simplified expression to get the final value.
8For continuity, compare the limit with f(a). For piecewise functions, check LHL and RHL separately.

Quickest way: Substitute, then match the option

When to use it: Use this in the MCQ paper when a limit looks long. It saves time and works for most Foundation-level questions.

Put in x = a. If the answer is a clean number, mark it and move on.
If you get 0/0 and the form is (xⁿ − aⁿ)/(x − a), write n·aⁿ⁻¹ straight away.
For a polynomial ratio as x→∞, compare only the highest powers. If the degrees are equal, the limit is the ratio of the leading coefficients.
If the degree of the numerator is lower, the limit is 0.
Check that your answer fits exactly one option before spending more time.
Skip a question if it needs more than about two minutes. Wrong answers cost 0.25 marks.

Common mistakes in Limits and Continuity Basics

Treating 0/0 as 0 or as 1.
Students think a number divided by itself is 1, or that 0 on top means 0.
Fix: 0/0 is an indeterminate form. Simplify the expression first, then substitute again.
Confusing the limit with the value of the function at the point.
The two are equal for simple functions, so students assume they always are.
Fix: Remember that the limit describes nearby values. Continuity requires the limit to equal f(a).
Forgetting to check both the left-hand and right-hand limits for piecewise functions.
Students substitute into one branch only.
Fix: Use the branch for x < a for LHL and the branch for x > a for RHL. Compare the two.
Using (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹ when the denominator does not match.
Students memorise the result without checking the form.
Fix: The denominator must be exactly x − a and x must approach the same a. Otherwise factorise directly.
Applying lim (eˣ − 1)/x = 1 when x does not tend to 0.
Students match the pattern but ignore the condition.
Fix: Check that the variable inside tends to 0 and that the same expression sits in the denominator.
At infinity, cancelling the wrong terms or treating ∞/∞ as 1.
Infinity is treated as a number.
Fix: Divide each term by the highest power of x, then let 1/x terms go to 0.

Worked examples

Example 1

lim (x→3) (x² − 9) ÷ (x − 3) equals: (A) 0 (B) 3 (C) 6 (D) 9

Show the solution

Substituting x = 3 gives 0/0, so simplify.
Factorise: x² − 9 = (x − 3)(x + 3).
Cancel (x − 3), which is allowed as x ≠ 3 in a limit.
The expression becomes x + 3.
Substitute x = 3: 3 + 3 = 6.

Answer: (C) 6

Example 2

lim (x→∞) (3x² + 5x) ÷ (2x² − 7) equals: (A) 0 (B) 3/2 (C) 5/7 (D) ∞

Show the solution

Both numerator and denominator have degree 2.
Divide every term by x²: (3 + 5/x) ÷ (2 − 7/x²).
As x→∞, 5/x → 0 and 7/x² → 0.
The limit is 3 ÷ 2 = 3/2.

Answer: (B) 3/2

Example 3

Let f(x) = 2x + 1 for x ≤ 2, and f(x) = k·x − 1 for x > 2. For f to be continuous at x = 2, k equals: (A) 1 (B) 2 (C) 3 (D) 4

Show the solution

LHL at x = 2: 2(2) + 1 = 5. Also f(2) = 5.
RHL at x = 2: k(2) − 1 = 2k − 1.
Continuity needs RHL = LHL = f(2).
So 2k − 1 = 5, which gives 2k = 6.
k = 3.

Answer: (C) 3

Exam tips

Always try direct substitution first. Many MCQs are solved in under 20 seconds this way.
Learn the n·aⁿ⁻¹ result and the three e-based standard limits. They are the most tested forms.
For continuity questions on piecewise functions, equate LHL, RHL and f(a) and solve for the unknown constant.
Check each option against your answer. Do not guess blindly, since a wrong answer loses 0.25 marks.
Practise polynomial ratios at infinity until you can read the answer from the leading terms.

Practice questions from Differential and Integral Calculus

Limits and Continuity Basics: frequently asked questions

What is the difference between limit and continuity?

A limit is the value f(x) approaches near a point. Continuity needs more: f(a) must exist and equal that limit. A function can have a limit at a point and still be discontinuous there if f(a) is missing or different.

How do I solve limits in Business Mathematics quickly?

Substitute first. If you get 0/0, factorise and cancel, or use a standard limit. For x→∞, divide by the highest power of x.

Which standard limit formulas should I remember for CA Foundation?

Remember lim (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹, lim (eˣ − 1)/x = 1, lim log(1 + x)/x = 1 and lim (1 + x)^(1/x) = e as x→0. Also remember that 1/xⁿ tends to 0 as x→∞.

When does a limit not exist?

A limit does not exist when the left-hand and right-hand limits are different. It also fails to exist as a finite number when the function grows without bound near the point.