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.

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

Continuity test at a point

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

Worked example

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

continuous at

x=1

Practice this conceptself-check · 4 quick reps

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]

Continuity needs the limit to EQUAL the value

A function can have a perfectly good limit at

a

and still be discontinuous — if

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

. All three of LHL, RHL, and

f(a)

must coincide. "The limit exists" alone is not continuity.

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.

Practice this conceptself-check · 4 quick reps

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

Practice this conceptself-check · 4 quick reps

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]

Removable vs jump — the limit's existence is the divider

A removable discontinuity has a single limit (LHL

=

RHL) that just misses

f(a)

— it can be patched. A jump has LHL

\neq

RHL (both finite), which can never be patched.

\lfloor x\rfloor

at integers is a jump, not removable.

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

Practice this conceptself-check · 4 quick reps

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

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

The definition of continuity
Continuity test at a point
$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$

Watch out for (2)

Continuity needs the limit to EQUAL the value
→ The definition of continuity
Removable vs jump — the limit's existence is the divider
→ Types of discontinuity (removable, jump, oscillatory)

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