10 min readConcept sheet1-mark MCQ and assertion-reason; a 2 or 3-mark 'find k' or 'find a and b' question; also a JEE favourite

Continuity of a function

f is continuous at x = c when the left-hand limit, the right-hand limit and f(c) are all equal. Its graph has no break there.

Continuity at x = clim (x → c⁻) f(x) = lim (x → c⁺) f(x) = f(c)

Continuous functionA function that is continuous at every point of its domain

Always continuousPolynomials, |x|, sin x and cos x everywhere; rational functions wherever the denominator is not 0

Algebra of continuityIf f and g are continuous at c, so are f + g, f − g, fg, and f/g when g(c) ≠ 0

What continuity means

Roughly, f is continuous at x = c if you can draw its graph through x = c without lifting the pen. Exactly, three things must agree. The left-hand limit (LHL) is the value f(x) approaches as x comes to c from values less than c. The right-hand limit (RHL) is the same from values greater than c. The third is the value f(c). If any one of them differs or does not exist, f is discontinuous at c.

How to test continuity at x = c

Use this for a function given by different formulas on each side of c.

1
Find f(c) from the formula that includes x = c.
2
Find the LHL: use the formula for x < c and let x → c.
3
Find the RHL: use the formula for x > c and let x → c.
4★
If LHL = RHL = f(c), f is continuous at c. Otherwise it is discontinuous at c.

★ key step

Example 1: test a point

Problem

Is f(x) = {x² + 1, if x ≤ 1; 3 − x, if x > 1} continuous at x = 1?

Working

f(1) = 1² + 1 = 2
LHL = lim (x → 1⁻) (x² + 1) = 2
RHL = lim (x → 1⁺) (3 − x) = 2
LHL = RHL = f(1) = 2

Answer

Yes, f is continuous at x = 1.

Example 2: find k

Problem

Find k so that f(x) = {kx + 3, if x ≤ 2; 2x² − 1, if x > 2} is continuous at x = 2.

Working

f(2) = 2k + 3, and LHL = lim (x → 2⁻) (kx + 3) = 2k + 3
RHL = lim (x → 2⁺) (2x² − 1) = 8 − 1 = 7
For continuity: 2k + 3 = 7
2k = 4, so k = 2

Answer

k = 2

Example 3: a limit with sine

Problem

Find k so that f(x) = {(sin 3x)/x, if x ≠ 0; k, if x = 0} is continuous at x = 0.

Working

Use the standard limit lim (θ → 0) (sin θ)/θ = 1.
lim (x → 0) (sin 3x)/x = lim (x → 0) 3 × (sin 3x)/(3x) = 3 × 1 = 3
The same formula holds on both sides of 0, so LHL = RHL = 3.
For continuity, f(0) = k must equal 3.

Answer

k = 3

Remember

The greatest integer function [x] is discontinuous at every integer: it jumps by 1 there.
f(x) = 1/x is a continuous function: x = 0 is not in its domain, so continuity is not tested there.
Sums, differences, products and quotients (with a non-zero denominator) of continuous functions are continuous.
A composite of continuous functions is continuous, for example sin(x²) and |x − 3|.
To find two unknowns a and b, write one continuity equation at each break point.

In real life

Body temperature on a fever chart changes continuously: to go from 98°F to 102°F it passes through every value in between. A parking fee that jumps at each hour mark is discontinuous there.

Exam angle

1 marks: “At which points is the greatest integer function f(x) = [x] discontinuous?” At every integer. At x = n, the LHL is n − 1 but the RHL and f(n) are n.

2 marks: “Find k so that f(x) = {(1 − cos 4x)/(8x²), if x ≠ 0; k, if x = 0} is continuous at x = 0.” 1 − cos 4x = 2 sin²2x. So the limit is lim (2 sin²2x)/(8x²) = (1/4) × [lim (sin 2x)/x]² = (1/4) × 2² = 1. So k = 1.

3 marks: “Find a and b if f(x) = {3, if x ≤ 1; ax + b, if 1 < x < 4; 9, if x ≥ 4} is continuous everywhere.” At x = 1: a + b = 3. At x = 4: 4a + b = 9. Subtracting, 3a = 6, so a = 2 and b = 1.

Remember it

Left, right and value: all three must match.