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

Continuity of a Function

Updated 1 October 2026 · Fact-checked

A function is continuous at x = a if f(a) is defined, the limit as x approaches a exists, and the limit equals f(a). To solve, find the left and right limits, compare them with f(a), and if a constant is unknown, set all three equal and solve for it.

Understand Continuity of a Function

Think of continuity as a graph you can draw without lifting your pen. If you must lift the pen at a point, the function is discontinuous there.

Maths turns this idea into three checks at a point x = a. First, f(a) must exist. Second, the limit of f(x) as x → a must exist, which means the left-hand limit (x approaches a from below) equals the right-hand limit (x approaches a from above). Third, the limit must equal f(a). Fail any one check and the function is discontinuous at a.

Discontinuities come in kinds. In a removable discontinuity, the limit exists but either f(a) is missing or f(a) is a different value. You could repair it by redefining f(a). In a jump discontinuity, the left and right limits both exist but are different, so the graph jumps. In an infinite discontinuity, the function blows up near a, like 1/x at x = 0.

Useful facts: polynomials are continuous everywhere. A rational function (one polynomial divided by another) is continuous at every point where the denominator is not zero. Piecewise functions usually need checking only at the point where the rule changes.

In exam questions, you are often told a function is continuous and asked for an unknown constant such as k. You then simply force the three values to be equal and solve a small equation.

Key formulas to remember

Condition for continuity at x = a: lim (x→a) f(x) = f(a); This needs f(a) defined and the limit to exist, so LHL = RHL = f(a).
One-sided limits: LHL = lim (x→a⁻) f(x), RHL = lim (x→a⁺) f(x); The limit exists only when LHL = RHL (both finite).
Removable discontinuity: lim (x→a) f(x) exists, but f(a) is undefined or ≠ the limit; Can be removed by defining f(a) equal to the limit.
Jump discontinuity: LHL ≠ RHL (both finite); The limit does not exist. It cannot be removed.
Finding an unknown constant: LHL = RHL = f(a); Put the boundary value into each piece, equate, and solve.
Standard factor cancellation: (x² − a²) ÷ (x − a) = x + a, for x ≠ a; The most common way to get a limit of a 0/0 form.

How to solve Continuity of a Function questions

Use this method for any question that asks whether a function is continuous at a point, what type of discontinuity it has, or what constant makes it continuous.

1Identify the point x = a to test. For a piecewise function it is usually where the rule changes.
2Find f(a) from the piece that includes x = a. If no value is given, f(a) is undefined.
3Find the left-hand limit using the rule for x < a. Substitute directly, or factor and cancel if you get 0/0.
4Find the right-hand limit using the rule for x > a in the same way.
5Compare. If LHL ≠ RHL, it is a jump discontinuity. If LHL = RHL but not equal to f(a) or f(a) is missing, it is removable. If a limit is infinite, it is infinite discontinuity.
6If a constant like k is unknown, write LHL = RHL = f(a), pick the equation that contains k, and solve it.
7Check your answer by substituting k back and confirming all three values match.

Quickest way: Plug in the boundary, equate, solve

When to use it: Use for MCQs that ask for k (or a and b) so the function is continuous at a point.

Ignore the full theory. Substitute the boundary value into each piece that is continuous there, as polynomial pieces need no limit work.
If a piece gives 0/0, factor the top and cancel the common factor, then substitute.
Set the two values equal and solve the one-line equation.
Plug each option into the equation to confirm only one fits. This is a quick check for four-option MCQs.
If two constants are unknown, you will get two equations from two boundary points. Solve them together.
Skip a question if it needs a long trigonometric limit you do not recall. A wrong answer costs 0.25 marks.

Common mistakes in Continuity of a Function

Checking only that the limit exists and ignoring f(a).
Students remember limits well and forget that the function value must also match.
Fix: Always write all three values: LHL, RHL and f(a). Continuity needs all three equal.
Cancelling (x − a) and then saying f(a) equals the cancelled expression.
The cancelled form hides the fact that the original function is undefined at x = a.
Fix: Use the cancelled form only to find the limit. f(a) comes from the given definition at a.
Using the wrong piece for a one-sided limit.
Students mix up which rule applies for x < a and which for x > a.
Fix: Mark the rule for x < a for the LHL and the rule for x > a for the RHL. The piece with ≤ or ≥ also gives f(a).
Calling every discontinuity removable.
Students notice a hole in the graph and do not compare LHL and RHL.
Fix: If LHL ≠ RHL, it is a jump discontinuity and cannot be removed, even if f(a) is defined.
Writing that a rational function is discontinuous everywhere.
Students confuse one bad point with the whole function.
Fix: A rational function is discontinuous only where the denominator is zero. Elsewhere it is continuous.
Solving for k with an arithmetic slip in the equation.
Rushing under time pressure, especially with negative signs.
Fix: Substitute your k back into both sides to confirm they match before marking an option.

Worked examples

Example 1

Let f(x) = (x² − 9) ÷ (x − 3) for x ≠ 3, and f(3) = k. If f is continuous at x = 3, what is k? Options: (a) 3 (b) 6 (c) 9 (d) 0

Show the solution

For x ≠ 3, x² − 9 = (x − 3)(x + 3), so f(x) = x + 3.
LHL = RHL = lim (x→3) (x + 3) = 6.
For continuity, f(3) must equal the limit, so k = 6.

Answer: (b) 6

Example 2

Let f(x) = 2x + 1 for x < 2 and f(x) = kx − 1 for x ≥ 2. If f is continuous at x = 2, what is k? Options: (a) 2 (b) 3 (c) 5 (d) 6

Show the solution

LHL = lim (x→2⁻) (2x + 1) = 2(2) + 1 = 5.
f(2) = k(2) − 1 = 2k − 1, and RHL is also 2k − 1.
Set LHL = f(2): 2k − 1 = 5, so 2k = 6 and k = 3.
Check: with k = 3, f(2) = 6 − 1 = 5, matching the LHL.

Answer: (b) 3

Example 3

Let f(x) = |x| ÷ x for x ≠ 0 and f(0) = 1. At x = 0, the function has: (a) a continuous point (b) a removable discontinuity (c) a jump discontinuity (d) an infinite discontinuity

Show the solution

For x < 0, |x| = −x, so f(x) = −x ÷ x = −1. Hence LHL = −1.
For x > 0, |x| = x, so f(x) = 1. Hence RHL = 1.
LHL ≠ RHL, and both are finite, so the limit does not exist.
The graph jumps from −1 to 1. Changing f(0) cannot fix this, so it is not removable.

Answer: (c) a jump discontinuity

Exam tips

For k-type MCQs, only the boundary matters. Substitute the boundary into each piece and equate. This takes under a minute.
Watch which side holds the equality sign (≤ or ≥). It tells you which piece gives f(a).
Before doing heavy algebra, test whether direct substitution gives 0/0. Only then factor and cancel.
For type-of-discontinuity questions, compute LHL and RHL first. If they differ, answer jump immediately.
Use the options to verify. Substitute each candidate k into the equation, and use this to eliminate wrong choices quickly.

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

Continuity of a Function: frequently asked questions

How do I check continuity at a point?

Check three things at x = a: f(a) is defined, the limit exists (LHL = RHL), and the limit equals f(a). If all three hold, the function is continuous at a. If any one fails, it is discontinuous.

How do I find k for continuity?

Find the limit from the side that does not contain k, and the value of f(a) or the other limit that contains k. Equate them and solve for k. Substitute back to confirm.

What is the difference between removable and jump discontinuity?

In a removable discontinuity the limit exists but f(a) is missing or different, so redefining f(a) fixes it. In a jump discontinuity the left and right limits are different, so no single value of f(a) can fix it.

Is every polynomial continuous?

Yes. A polynomial is continuous at every real number. A rational function is continuous everywhere except where its denominator is zero.