Is the absolute value function continuous at zero?

Yes. Both one-sided limits equal zero and the function value at zero is zero. It is continuous everywhere; it is not differentiable at zero because of the corner, but continuity and differentiability are different properties.

How does the Intermediate Value Theorem help solve equations?

It is a root-existence tool. If a continuous function has opposite signs at the endpoints of an interval, it must have at least one zero inside. Combined with monotonicity, this often gives uniqueness of the root.

Continuity of Function

Q: What is the difference between 'limit exists' and 'continuous'?

The limit at a exists when left-hand and right-hand limits agree and are finite. Continuity requires additionally that this common value equals f(a). Continuity is limit existence plus agreement with the function value.

Q: Are polynomials continuous everywhere?

Yes. Every polynomial is continuous on all real numbers because it is built from constants and x using operations that preserve continuity. Rational functions are continuous everywhere except at zeros of the denominator.

Q: How do I classify a discontinuity?

Compute the left-hand and right-hand limits. If both are finite and equal but the function value differs or is undefined, it is removable. If both are finite but unequal, it is a jump. If at least one is infinite or does not exist due to oscillation, it is essential.

Q: Can a function be continuous at only one point?

Yes. The classic example is the function that equals x on the rationals and minus x on the irrationals; it is continuous only at zero.

Q: What if I need to check continuity at infinitely many points?

Identify the general form. For step or fractional-part functions, discontinuities occur at every integer with identical analysis. Check one representative point and note it applies to all.

Q: Does continuity imply differentiability?

No. Continuity is necessary but not sufficient for differentiability. The absolute value function is continuous but not differentiable at zero. Some functions are continuous everywhere but differentiable nowhere.

MathsLimits, Continuity And DifferentiabilityFor JEE aspirants

A function $f (x)$ is continuous at $x = a$ if the graph has no break, hole, or jump at that point. The precise condition is a three-way equality: $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$ , all finite. So continuity requires the LHL and RHL to exist, be equal, and match the function's actual value at $a$ . When this fails, the discontinuity is classified as removable, jump, or essential. Standard functions - polynomials, $sin, cos, e^{x}, ln$ (in its domain) - are continuous throughout their domains. Continuity is the bridge from limits to differentiability and underlies the Intermediate Value Theorem and Extreme Value Theorem central to JEE calculus.

Key Conditions - Quick Reference

Continuity at a point: $f$ is continuous at $x = a$ iff $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$ , all finite.
Discontinuity: if any one of the three quantities does not exist, or all three exist but are not all equal, $f$ is discontinuous at $a$ .
Open interval: $f$ is continuous on $(a, b)$ if it is continuous at every point of $(a, b)$ .
Closed interval: $f$ is continuous on $[a, b]$ if it is continuous on $(a, b)$ , $lim_{x \to a^{+}} f (x) = f (a)$ , and $lim_{x \to b^{-}} f (x) = f (b)$ .
Algebra: if $f, g$ are continuous at $a$ , so are $f \pm g$ , $f \cdot g$ , and $f / g$ (last one requires $g (a) \neq = 0$ ).
Composition: if $g$ is continuous at $a$ and $f$ is continuous at $g (a)$ , then $f \circ g$ is continuous at $a$ .
Intermediate Value Theorem: if $f$ is continuous on $[a, b]$ and $λ$ is between $f (a)$ and $f (b)$ , there exists $c \in (a, b)$ with $f (c) = λ$ .

1. Definition of Continuity

A function

f (x)

is continuous at $x = a$ if

x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = f (a),

with all three quantities finite. If continuity fails at

a

, then

f

is discontinuous at $a$ .

Equivalent short forms of the same condition:

LHL $=$ RHL $=$ value of the function at $a$ .
$lim_{x \to a} f (x) = f (a)$ .

Isolated points. We do not talk about continuity at points that cannot be approached from either side. For instance,

f (x) = α - x + x - α

has only

x = α

in its domain; continuity at

α

is not discussed. Standard classes of functions - polynomials, trigonometric, exponential, and logarithmic functions - are continuous throughout their domains.

2. Reasons for Discontinuity

A function fails to be continuous at $a$ for one of the following reasons.

2.1 One or more of LHL, RHL, or $f (a)$ does not exist

Example: $f (x) = \frac{1}{x}$ at $x = 0$ . Here LHL $= - \infty$ , RHL $= + \infty$ , and $f (0)$ is undefined. The function is discontinuous at $0$ because all three quantities fail simultaneously.

Example (removable): $f (x) = \frac{x ^{2} - 1}{x - 1}$ around $x = 1$ . Here LHL $=$ RHL $= 2$ but $f (1)$ is undefined. The graph has a "hole" at $x = 1$ .

2.2 All three quantities exist but at least one pair disagrees

Example: $f (x) = [x]$ (greatest integer) at any integer $I$ . LHL $= I - 1$ , RHL $= I$ , $f (I) = I$ . Since LHL $\neq =$ RHL, $f$ is discontinuous at every integer.

Example: $f (x) = {x}$ (fractional part) at any integer $I$ . LHL $= 1$ , RHL $= 0$ , $f (I) = 0$ . LHL $\neq =$ RHL, so discontinuous at every integer.

Example: $f (x) = {1, x \in / Z 0, x \in Z$ at any integer $I$ . LHL $= 1$ , RHL $= 1$ , but $f (I) = 0$ . The limit exists and equals $1$ , but $f (I) \neq = 1$ , so the function is discontinuous.

Example: $f (x) = {∣ x ∣/ x, x \neq = 0 0, x = 0$ at $x = 0$ . LHL $= - 1$ , RHL $= 1$ , $f (0) = 0$ . All three are different, so discontinuous.

Figure 1: Three archetypes of discontinuity. Removable: the limit exists but

f (a)

is missing or different. Jump: LHL and RHL exist but disagree. Essential: at least one one-sided limit is infinite or fails to exist.

3. Types of Discontinuity

Type	Characterisation	Fixable?
Removable	LHL $=$ RHL (finite), but $f (a)$ is undefined or $\neq =$ that common value	Yes, by redefining $f (a)$
Jump	LHL and RHL both exist and are finite, but LHL $\neq =$ RHL	No
Essential (or oscillatory)	At least one of LHL, RHL is $\pm \infty$ or does not exist (oscillation)	No

Oscillatory example:

f (x) = sin (1/ x)

near

x = 0

. As

x \to 0

, the argument

1/ x

swings between

- \infty

and

\infty

, and

sin

oscillates between

- 1

and

1

infinitely often. Neither LHL nor RHL exists, so the discontinuity at

0

is essential.

4. Solved Examples: Checking Continuity

Solved Example 1

Discuss the continuity of

f (x)

x = \frac{1}{2}

, where

f (x) = ⎩ ⎨ ⎧ \frac{1}{2} - x, 0 \leq x < \frac{1}{2} 1, x = \frac{1}{2} \frac{3}{2} - x, \frac{1}{2} < x \leq 1.

Solution:

LHL: $lim_{x \to 1/ 2^{-}} f (x) = \frac{1}{2} - \frac{1}{2} = 0$ .

RHL: $lim_{x \to 1/ 2^{+}} f (x) = \frac{3}{2} - \frac{1}{2} = 1$ .

$f (1/2) = 1$ .

Since LHL $= 0 \neq = 1 =$ RHL, the function is discontinuous at $x = 1/2$ . This is a jump discontinuity.

Solved Example 2

Discuss the continuity of

f (x) = {2 - x, x < 2 2 + x, x \geq 2

x = 2

Solution:

LHL: $lim_{x \to 2^{-}} (2 - x) = 0$ .

RHL: $lim_{x \to 2^{+}} (2 + x) = 4$ .

Since LHL $\neq =$ RHL, $f$ is discontinuous at $x = 2$ . Jump discontinuity.

Solved Example 3

f (x) = ⎩ ⎨ ⎧ 2 x + 3, x < 0 0, x = 0 x^{2} + 3, x > 0,

discuss continuity at

x = 0

Solution:

LHL: $lim_{x \to 0^{-}} (2 x + 3) = 3$ .

RHL: $lim_{x \to 0^{+}} (x^{2} + 3) = 3$ .

$f (0) = 0$ .

LHL $=$ RHL $= 3$ , but $f (0) = 0 \neq = 3$ . This is a removable discontinuity: redefining $f (0) = 3$ would make the function continuous.

5. Continuity on an Interval

5.1 Open interval

f

is continuous on the open interval

(a, b)

if it is continuous at every point of

(a, b)

5.2 Closed interval

f

is continuous on the closed interval

[a, b]

if:

$f$ is continuous on $(a, b)$ ;
$lim_{x \to a^{+}} f (x) = f (a)$ (right-continuous at the left endpoint);
$lim_{x \to b^{-}} f (x) = f (b)$ (left-continuous at the right endpoint).

5.3 Geometrical meaning

A function is continuous on an interval if and only if you can draw its graph over that interval without lifting the pen. On an open interval, no breaks anywhere. On a closed interval, no breaks even at the endpoints.

Examples of intervals of continuity.

sin x

and

cos x

are continuous on all of

R

tan x

is discontinuous at

x = (2 n + 1) \frac{π}{2}

and continuous everywhere else.

ln x

is continuous on

(0, \infty)

x

is continuous on

[0, \infty)

, with right-continuity at

0

6. Continuity as a Formula-Fitting Problem

A classic JEE problem: a piecewise function contains an unknown constant, and we must choose the constant to make the function continuous. Match LHL, RHL, and $f (a)$ at each boundary point.

Solved Example 4

f (x) = {(tan (\frac{π}{4} + x))^{1/ x}, x \neq = 0 k, x = 0.

Find

k

for which

f

is continuous at

x = 0

Solution:

Compute $lim_{x \to 0} f (x)$ . Substituting $x = 0$ inside the base gives $tan (π /4) = 1$ , and $1/ x \to \infty$ , so this is the form $1^{\infty}$ .

x \to 0 lim (tan (\frac{π}{4} + x))^{1/ x} = x \to 0 lim (\frac{1 + tan x}{1 - tan x})^{1/ x} .

Apply the $1^{\infty}$ formula:

exp (x \to 0 lim \frac{1}{x} (\frac{1 + tan x}{1 - tan x} - 1)) = exp (x \to 0 lim \frac{2 tan x}{x ( 1 - tan x )}) = exp (2) = e^{2} .

For continuity, $k = f (0) = lim_{x \to 0} f (x) = e^{2}$ .

7. Algebra of Continuous Functions

If $f$ and $g$ are continuous at $x = a$ , then

$c f (x)$ is continuous at $a$ for any constant $c$ .
$f (x) \pm g (x)$ is continuous at $a$ .
$f (x) \cdot g (x)$ is continuous at $a$ .
$\frac{f ( x )}{g ( x )}$ is continuous at $a$ , provided $g (a) \neq = 0$ .

Mixed continuity - surprising combinations:

If $f$ is discontinuous at $a$ and $g$ is discontinuous at $a$ , then $f + g$ may still be continuous. Example: $f (x) = [x]$ , $g (x) = {x}$ , both discontinuous at every integer, but $(f + g) (x) = x$ is continuous everywhere.
If $f$ is continuous and $g$ is discontinuous at $a$ , then $f \cdot g$ can be continuous. Example: $f (x) = x$ , $g (x) = sin (1/ x)$ for $x \neq = 0$ , $g (0) = 0$ . Then $f \cdot g \to 0$ as $x \to 0$ , so continuous at $0$ .
If both are discontinuous, $f \cdot g$ can be continuous. Example: $f (x) = sgn (x)$ (with $f (0)$ any value), $g (x) = - sgn (x)$ ; then $f g = - 1$ except at $0$ .

8. Continuity of Composite Functions

g

is continuous at

c

and

f

is continuous at

g (c)

, then

(f \circ g) (x) = f (g (x))

is continuous at

c

To locate discontinuities of a composite, we find where each layer breaks: either $g$ is discontinuous at $c$ , or $g$ is continuous at $c$ but $f$ is discontinuous at $g (c)$ .

Solved Example 5

Find the points of discontinuity of

y = \frac{1}{u ^{2} + u - 2}

, where

u = \frac{1}{x - 1}

Solution:

Inner layer: $u = \frac{1}{x - 1}$ is discontinuous at $x = 1$ .

Outer layer: $y = \frac{1}{u ^{2} + u - 2} = \frac{1}{( u + 2 ) ( u - 1 )}$ is discontinuous where the denominator vanishes: $u = - 2$ or $u = 1$ .

Convert these $u$ -values back to $x$ :

$u = - 2 ⟹ \frac{1}{x - 1} = - 2 ⟹ x = \frac{1}{2}$ .

$u = 1 ⟹ \frac{1}{x - 1} = 1 ⟹ x = 2$ .

Combining, the composite is discontinuous at $x = \frac{1}{2}, 1, 2$ .

9. Intermediate Value Theorem

Intermediate Value Theorem (IVT). If

f

is continuous on

[a, b]

and

λ

is any number between

f (a)

and

f (b)

, then there exists at least one

c \in (a, b)

such that

f (c) = λ

Special case: if $f (a)$ and $f (b)$ have opposite signs, there exists $c \in (a, b)$ with $f (c) = 0$ . This is the fixed-point / root-existence tool used throughout JEE.

Figure 2: For a continuous

f

[a, b]

, any horizontal line

y = λ

with

λ

between

f (a)

and

f (b)

must cross the graph at some point

c

Solved Example 6

Let

f : [0, 1] \to [1, e]

be a continuous function. Prove that

f (c) = e^{c}

for some

c \in [0, 1]

Solution:

Consider $h (x) = f (x) - e^{x}$ , which is continuous on $[0, 1]$ as a difference of continuous functions.

At the endpoints:

$h (0) = f (0) - e^{0} = f (0) - 1 \geq 0$ (since $f (0) \in [1, e]$ , so $f (0) \geq 1$ ).

$h (1) = f (1) - e^{1} = f (1) - e \leq 0$ (since $f (1) \in [1, e]$ , so $f (1) \leq e$ ).

Since $h$ is continuous and changes sign (or touches zero) between $0$ and $1$ , by IVT there exists $c \in [0, 1]$ with $h (c) = 0$ , that is, $f (c) = e^{c}$ .

10. Extreme Value Theorem

Extreme Value Theorem (EVT). If

f

is continuous on a closed interval

[a, b]

, then

f

attains its absolute maximum and absolute minimum on

[a, b]

. That is, there exist

x_{1}, x_{2} \in [a, b]

such that

f (x_{1}) \leq f (x) \leq f (x_{2}) for all x \in [a, b] .

Closedness is essential: on $(0, 1)$ , the function $f (x) = x$ has no maximum. Continuity is essential: a jumping function on $[0, 1]$ may skip over its intended max.

Solved Example 7

Let

f : (0, 1) \to [0, 1]

be a continuous surjective function. Show that

f

cannot be one-to-one.

Solution:

Suppose for contradiction that $f$ is one-to-one. A continuous one-to-one function on an interval must be strictly monotonic, so its range would be an open interval (since $(0, 1)$ is open and $f$ is monotonic). But the range is $[0, 1]$ , which is closed - a contradiction. Therefore $f$ must be many-to-one.

Common Mistakes to Avoid

Watch out

Checking only $lim_{x \to a} f (x) = f (a)$ without confirming the limit exists. The limit itself requires LHL $=$ RHL. All three checks are needed at a boundary point of a piecewise function.
Confusing "limit exists" with "continuous." The limit at $a$ can exist while $f (a)$ is undefined or different; this is a removable discontinuity, not continuity.
Assuming that a sum of two discontinuous functions is discontinuous. As shown, $[x] + {x} = x$ is continuous everywhere.
Applying IVT without checking continuity on the whole closed interval. IVT fails if the function has even one discontinuity inside $[a, b]$ .
Applying EVT on an open interval. The Extreme Value Theorem requires a closed and bounded interval; on open intervals the theorem is silent.
Ignoring endpoint conditions in closed-interval continuity. Continuity on $[a, b]$ requires right-continuity at $a$ and left-continuity at $b$ .
Assuming isolated points are continuous or discontinuous. We simply do not discuss continuity at isolated points; the definition requires approaching the point from at least one side.

Frequently Asked Questions

Q1. What is the difference between "limit exists" and "continuous"?

The limit at $a$ exists when LHL $=$ RHL, both finite. Continuity requires additionally that this common value equals $f (a)$ . So continuity is limit existence plus agreement with the actual function value.

Q2. Are polynomials continuous everywhere?

Yes. Every polynomial is continuous on all of $R$ because it is built from constants and $x$ by sums, differences, and products - all operations that preserve continuity. Rational functions (ratios of polynomials) are continuous everywhere except at zeros of the denominator.

Q3. How do I classify a discontinuity?

Compute LHL and RHL. If both are finite and equal but $f (a)$ differs or is undefined: removable. If both are finite but unequal: jump. If at least one is $\pm \infty$ or does not exist (oscillation): essential.

Q4. Can a function be continuous at only one point?

Yes. The classic example is $f (x) = {x, x \in Q - x, x \in / Q$ , which is continuous only at $x = 0$ . Both pieces agree there and give the value $0$ .

Q5. Is $∣ x ∣$ continuous at $x = 0$ ?

Yes. $lim_{x \to 0^{-}} ∣ x ∣ = 0 = lim_{x \to 0^{+}} ∣ x ∣$ and $∣0∣ = 0$ ; all three agree. $∣ x ∣$ is continuous everywhere. It is not differentiable at $0$ (a corner), but continuity and differentiability are different properties.

Q6. How does IVT help me solve equations?

IVT is a root-existence tool. If $f$ is continuous on $[a, b]$ and $f (a)$ and $f (b)$ have opposite signs, then $f (x) = 0$ has at least one solution in $(a, b)$ . Combined with monotonicity, this often gives uniqueness of the root.

Q7. What if I need to check continuity at infinitely many points?

Identify the general form. For $[x]$ or ${x}$ , discontinuities occur at every integer, and by symmetry the analysis at $x = I$ is identical for every integer $I$ . Check one representative point and note it applies to all integers.

Q8. Does continuity imply differentiability?

No. Continuity is a necessary condition for differentiability but not sufficient. $∣ x ∣$ is continuous at $0$ but not differentiable there; $3 x$ is continuous at $0$ but has a vertical tangent. Some functions (like the Weierstrass function) are continuous everywhere but differentiable nowhere.

Previous year questions on Continuity of Function

18 questions from past papers, each with a step-by-step solution.

Show all 18 questions

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1. Definition of Continuity

2. Reasons for Discontinuity

2.1 One or more of LHL, RHL, or f(a) does not exist

2.2 All three quantities exist but at least one pair disagrees

3. Types of Discontinuity

4. Solved Examples: Checking Continuity

5. Continuity on an Interval

5.1 Open interval

5.2 Closed interval

5.3 Geometrical meaning

6. Continuity as a Formula-Fitting Problem

7. Algebra of Continuous Functions

8. Continuity of Composite Functions

9. Intermediate Value Theorem

10. Extreme Value Theorem

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between "limit exists" and "continuous"?

Q2. Are polynomials continuous everywhere?

Q3. How do I classify a discontinuity?

Q4. Can a function be continuous at only one point?

Q5. Is ∣x∣ continuous at x=0?

Q6. How does IVT help me solve equations?

Q7. What if I need to check continuity at infinitely many points?

Q8. Does continuity imply differentiability?

Previous year questions on Continuity of Function

Ready to master Limits, Continuity And Differentiability?

2.1 One or more of LHL, RHL, or $f (a)$ does not exist

Q5. Is $∣ x ∣$ continuous at $x = 0$ ?