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Continuity at a Point

A function f is continuous at x = a if f(a) is defined, lim(x -> a) f(x) exists, and lim(x -> a) f(x) = f(a).

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
A function f is continuous at x = a if f(a) is defined, lim(x → a) f(x) exists, and lim(x → a) f(x) = f(a).
2
Then show how to use it
Identify the point of possible discontinuity. Find the left-hand limit. Find the right-hand limit. Find f(a) using the rule that includes equality. Equate LHL, RHL, and f(a), then solve for the unknown constants.
3
Add one concrete example
For f(x) = {kx + 1, x <= 2; 3x - 1, x > 2}, continuity at x = 2 requires 2k + 1 = 3(2) - 1 = 5, so k = 2.
4
Avoid this incomplete answer
Using the expression for x > a to calculate f(a) when the function actually defines f(a) in the x <= a branch.

Definition

A function f is continuous at x = a if f(a) is defined, lim(x → a) f(x) exists, and lim(x → a) f(x) = f(a).

Example

For f(x) = {kx + 1, x <= 2; 3x - 1, x > 2}, continuity at x = 2 requires 2k + 1 = 3(2) - 1 = 5, so k = 2.

Rule to remember

Continuity at x = a: LHL = RHL = f(a). Algebra of continuous functions: if f and g are continuous at a, then f + g, f - g, fg, and f/g are continuous at a, provided g(a) != 0.

Memory hook

Continuity needs three matches: left value, right value, and actual value.

Examples and method

Worked example

Let f(x) = {x + 3, x <= 1; px + q, 1 < x < 3; 2x - 1, x >= 3}. For continuity at x = 1: 1 + 3 = p + q, so p + q = 4. For continuity at x = 3: 3p + q = 2(3) - 1 = 5. Subtracting, 2p = 1, so p = 1/2. Then q = 4 - 1/2 = 7/2. Hence p = 1/2 and q = 7/2.

Method to apply

Identify the point of possible discontinuity. Find the left-hand limit. Find the right-hand limit. Find f(a) using the rule that includes equality. Equate LHL, RHL, and f(a), then solve for the unknown constants.

Diagram support

A graph may help show a break, jump, or removable gap, but algebraic one-sided limit checking is enough for standard CBSE questions on this concept.

How CBSE asks it

Usually asked by giving a piecewise function and asking for unknown constants that make it continuous at one or two points.

Avoid common mistakes

Common confusion

Students often equate only the two formula pieces and forget to compare them with f(a), especially when the equality sign belongs to one side of a piecewise function.

Common wrong answer

Using the expression for x > a to calculate f(a) when the function actually defines f(a) in the x <= a branch.

Exam tip

At a joining point, first decide which expression gives f(a), then compare it with the left-hand and right-hand limits.

Quick check

If f(x) = x^2 for x < 1 and f(x) = ax + 2 for x >= 1 is continuous at x = 1, find a.

Left-hand limit = 1. Since f(1) = a + 2, continuity gives a + 2 = 1, so a = -1.

Answer writing and exam use

1-mark answer

A function f is continuous at x = a if f(a) is defined, lim(x → a) f(x) exists, and lim(x → a) f(x) = f(a).

2-mark answer

A function f is continuous at x = a if f(a) is defined, lim(x → a) f(x) exists, and lim(x → a) f(x) = f(a). Continuity at x = a: LHL = RHL = f(a). Algebra of continuous functions: if f and g are continuous at a, then f + g, f - g, fg, and f/g are continuous at a, provided g(a) != 0. For f(x) = {kx + 1, x <= 2; 3x - 1, x > 2}, continuity at x = 2 requires 2k + 1 = 3(2) - 1 = 5, so k = 2.

3-mark answer

Continuity means there is no break in the function at the point being tested. In exam questions, check the left-hand limit, right-hand limit, and actual value of the function. For piecewise functions, the point where the rule changes is the main point to test. Continuity at x = a: LHL = RHL = f(a). Algebra of continuous functions: if f and g are continuous at a, then f + g, f - g, fg, and f/g are continuous at a, provided g(a) != 0. Let f(x) = {x + 3, x <= 1; px + q, 1 < x < 3; 2x - 1, x >= 3}. For continuity at x = 1: 1 + 3 = p + q, so p + q = 4. For continuity at x = 3: 3p + q = 2(3) - 1 = 5. Subtracting, 2p = 1, so p = 1/2. Then q = 4 - 1/2 = 7/2. Hence p = 1/2 and q = 7/2. Usually asked by giving a piecewise function and asking for unknown constants that make it continuous at one or two points. Using the expression for x > a to calculate f(a) when the function actually defines f(a) in the x <= a branch.

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