NDA Maths · Limits & Continuity

Continuity & Differentiability

A function is continuous at a point when the left limit, the right limit, and the function's value all agree — and continuity is the necessary (not sufficient) condition for differentiability.

All Limits & Continuity notes NDA Maths strategy

Why this matters

Most continuity questions either ask you to fix a parameter so the pieces meet, or to classify a discontinuity. The recurring trap is the continuous-but-not-differentiable corner, and oscillatory functions like sin(1/x) that have no limit at all.

Concept 1 of 4

The definition of continuity

Intuition

Continuous at

a

means the graph has no break there: the value you approach equals the value you land on. Three things must all be equal — the left limit, the right limit, and

f(a)

Definition

$f$ is **continuous at $a$ ** iff $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$ (all three exist and are equal). Polynomials, $\sin$ , $\cos$ , $e^x$ are continuous everywhere; rational functions are continuous except where the denominator vanishes. A removable discontinuity (a 0/0 hole) is patched by defining $f(a)=\lim_{x\to a}f(x)$ .

Worked example

f(x)=\begin{cases}\dfrac{x^2-1}{x-1},&x\neq 1\\ 2,&x=1\end{cases}

continuous at

x=1

$\lim_{x\to1}\dfrac{x^2-1}{x-1}=\lim_{x\to1}(x+1)=2$ .
This equals $f(1)=2$ .

Answer:Yes — continuous at

x=1

Practice this conceptself-check · 4 quick reps

Try it yourself

f(x)=\dfrac{x^2-25}{x-5}

(

x\neq 5

) is made continuous at

x=5

. What value must

f(5)

take?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Continuity at $a$ needs which three equal?
2.
Are polynomials continuous everywhere?
3.
A removable discontinuity is patched by?
4.
Where can a rational function be discontinuous?

From the bank · past-year question

Example 1Limits & ContinuityEASY

f(x)=\frac{x^2-9}{x^2-2x-3}

x\ne3

is continuous at x=3, then?

[Q80 · Apr · 2018]

Drill 6 more on the definition of continuity

Concept 2 of 4

Finding parameters so f is continuous

Intuition

When a piecewise function carries unknown constants and is declared continuous, set the one-sided limits equal to each other and to

f

at every join. Each join gives one equation; solve the system for the unknowns.

Definition

At each join $x=c$ : impose $\lim_{x\to c^-}f=\lim_{x\to c^+}f=f(c)$ . With $k$ joins and $k$ unknowns you get $k$ equations — solve simultaneously. (Continuity needs only value-matching; differentiability would additionally need slope-matching.)

Worked example

Find

k

f(x)=\begin{cases}kx+1,&x\le 2\\ 3x-1,&x>2\end{cases}

is continuous.

Match at $x=2$ : $k(2)+1=3(2)-1\Rightarrow 2k+1=5$ .
$k=2$ .

Answer:

k=2

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

k

f(x)=\begin{cases}\dfrac{\sin x}{x},&x\neq 0\\ k,&x=0\end{cases}

is continuous at 0.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Each join contributes how many equations?
2.
Continuity matches values; differentiability also matches?
3.
$kx+1$ & $3x-1$ meet at $x=2$ : $k$ ?
4.
$\frac{\sin x}{x}$ patched at 0 needs $k=?$

From the bank · past-year question

Example 2Limits & ContinuityEASY

For what value of

k

is the function

f(x) = \begin{cases} 2x + \dfrac{1}{4}, & x < 0 \\ k, & x = 0 \\ \left(x + \dfrac{1}{2}\right)^{2}, & x > 0 \end{cases}

continuous ?

[Q98 · Sep · 2019]

Drill 10 more on finding parameters so f is continuous

Concept 3 of 4

Types of discontinuity (removable, jump, oscillatory)

Intuition

Discontinuities come in flavours: a removable hole (the limit exists but ≠ f(a)), a jump (LHL and RHL both exist but differ), and an oscillatory/essential break where no limit exists — the classic being sin(1/x) near 0. Greatest-integer functions create jump discontinuities at every integer.

Definition

Removable: $\lim_{x\to a}f$ exists but $\neq f(a)$ (or $f(a)$ undefined) — patchable.
Jump: LHL $\neq$ RHL, both finite (e.g. $\lfloor x\rfloor$ at integers).
Oscillatory/essential: no limit — $\sin\tfrac1x$ and $\sin\tfrac{1}{x^2}$ oscillate infinitely near 0.

Greatest-integer-built functions like $\lfloor x\rfloor^2-\lfloor x^2\rfloor$ are discontinuous at integers.

Worked example

Classify the discontinuity of

f(x)=\sin\tfrac1x

x=0

As $x\to0$ , $\tfrac1x\to\pm\infty$ and $\sin\tfrac1x$ oscillates between $-1$ and $1$ without settling.
No limit exists from either side.

Answer:Oscillatory (essential) discontinuity.

Practice this conceptself-check · 4 quick reps

Try it yourself

\lim_{x\to 0}\sin\tfrac{1}{x^2}

defined? Classify.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Removable discontinuity: limit exists but?
2.
Jump discontinuity: LHL and RHL are?
3.
Does $\lim_{x\to0}\sin\tfrac1x$ exist?
4.
$\lfloor x\rfloor$ has which discontinuity at integers?

From the bank · past-year question

Example 3Limits & ContinuityMODERATE

Consider the following statements in respect of the function

f(x)=\sin\!\left(\frac{1}{x^{2}}\right)

x\ne0

: 1. It is continuous at

x=0

, if

f(0)=0

. 2. It is continuous at

x=\frac{2}{\sqrt{\pi}}

. Which of the above statements is/are correct?

[Q78 · Sep · 2021]

Drill 4 more on types of discontinuity (removable, jump, oscillatory)

Concept 4 of 4

Continuity vs differentiability

Intuition

Differentiability is strictly stronger: differentiable ⇒ continuous, but not the reverse. The standard counter-examples are corners (

|x|

) and steps (

\lfloor x\rfloor

) — continuous (for the corner) yet not differentiable. Continuity is also preserved by sums, products, and composition of continuous functions.

Definition

**Differentiable at $a$ ⇒ continuous at $a$ ** (not conversely). $|x|$ is continuous at 0 but not differentiable (corner).
Closure: if $f,g$ are continuous at $a$ , so are $f\pm g$ , $fg$ , $f\circ g$ , and $f/g$ (where $g(a)\neq0$ ).
A product can be continuous even when a factor is awkward (e.g. $x\sin\tfrac1x\to 0$ by the squeeze).

Worked example

f(x)=|x|

continuous and differentiable at

x=0

Continuity: $\lim_{x\to0}|x|=0=f(0)$ — continuous.
Differentiability: left slope $-1$ , right slope $+1$ — a corner, not differentiable.

Answer:Continuous but not differentiable at 0.

Practice this conceptself-check · 4 quick reps

Try it yourself

f

is differentiable at

x=a

, is it continuous there?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Differentiable ⇒ ?
2.
Continuous ⇒ differentiable?
3.
$|x|$ at 0: continuous? differentiable?
4.
Is $f\circ g$ continuous if $f,g$ are?

From the bank · past-year question

Example 4Limits & ContinuityEASY

f(x)

is differentiable at

x=a

, then consider the following statements: I.

f(x)

is continuous at

x=a

. II.

\lim_{x\to a}f(x)=f(a)

. Which of the statements given above is/are correct?

[Q71 · Apr · 2026]

Drill 10 more on continuity vs differentiability

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Limits & ContinuityMODERATE

Consider the following: 1.

x + x^2

is continuous at

x = 0

. 2.

x + \cos\dfrac{1}{x}

is discontinuous at

x = 0

. 3.

x^2 + \cos\dfrac{1}{x}

is continuous at

x = 0

. Which of the above are correct?

[Q64 · Sep · 2017]

Example 2Limits & ContinuityEASY

for the items that follow: Let

f(x)=\begin{cases}ax(x-1), & x<1\\ x-1, & 1\leq x\leq3\\ px^2+qx+2, & x>3\end{cases}

. Given that

f(x)

is continuous for all x but not differentiable at x=1. Further

f'(x)

is continuous at x=3.

What is the value of q?

[Q96 · Apr · 2026]

Example 3Limits & ContinuityEASY

f(x)=\frac{\sin x}{x}

, where

x\in\mathbf{R}

, is to be continuous at

x=0

, then the value of the function at

x=0

[Q81 · Apr · 2020]

Example 4Limits & ContinuityMODERATE

Consider the following statements for

f(x)=e^{-|x|}

: 1. The function is continuous at

x=0

. 2. The function is differentiable at

x=0

. Which of the above statements is/are correct?

[Q91 · Apr · 2020]

Example 5Limits & ContinuityEASY

A function is defined as follows:

f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \neq 0 \\ 0, & x = 0 \end{cases}

. Which one of the following is correct in respect of the above function?

[Q62 · Sep · 2017]

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