Continuity (Layered)

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2026-09-12

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Layered chapter on continuity: the three-condition test, discontinuity taxonomy, IVT, question-pattern anatomy, and attempt-first checks. # Continuity > [!SUMMARY] One sitting > This chapter is designed for one focused session (roughly 45–60 minutes with the checks).

Continuity

One sitting

This chapter is designed for one focused session (roughly 45–60 minutes with the checks). By the end you will test continuity with a three-step ladder, classify every discontinuity papers ask about, apply the Intermediate Value Theorem to locate roots, and know exactly what to do next.

Where this chapter sits

Continuity is the payoff of limits: a function is continuous at a point when its limit equals its value there. Almost every evaluation technique from the Limits chapter returns here as "the function is continuous, so substitute directly". Derivatives (next chapter) are defined only where this property holds.

You need first:

Limits of Functions — two-sided limits, limit laws, factoring and rationalizing $\tfrac{0}{0}$ forms
Functions and their types — reading domain, pieces, and behavior near a point

This unlocks:

Derivatives — Definition and Rules — differentiability implies continuity (the converse fails; that failure is examinable)
Definite integrals — integrability rests on continuity over closed intervals

How to read the symbols

Symbol	Read it aloud as	What it means	Worth remembering
$f$ is continuous at $a$	"f is continuous at a"	the three conditions below all hold at $x = a$	a property of a point; "continuous function" means continuous at every point of its domain
Removable discontinuity	"removable"	limit exists but the value is missing or wrong	fixable by redefining one point — hence the name
Jump discontinuity	"jump"	left and right limits exist but disagree	the graph leaps; no redefinition of one point can fix it
Infinite discontinuity	"infinite" (verbally: "blow-up")	outputs grow without bound near $a$	we sometimes write $\to \infty$ , but no finite limit exists
$C[a, b]$	"C on a-b"	the set of functions continuous at every point of $[a, b]$	papers use this shorthand in IVT statements

Layer 1 — Intuition: the pen test, then the numbers

The folk version: a function is continuous if you can draw its graph without lifting the pen. Holes, jumps, and breaks each force a lift. The folk version becomes mathematics the moment you ask where the pen lifts and why — that is the whole chapter.

Compare what happens near

x = 1

for

f(x) = \dfrac{x^2 - 1}{x - 1}

versus near

x = 0

for

g(x) = \dfrac{\lvert x \rvert}{x}

$x$ near $1$	$f(x)$	$x$ near $0$	$g(x)$
$0.9$	$1.9$	$-0.1$	$-1$
$0.99$	$1.99$	$-0.01$	$-1$
$1.01$	$2.01$	$0.01$	$+1$
$1.1$	$2.1$	$0.1$	$+1$

f

trends to

2

from both sides (one hole at

x = 1

, since

f(1)

is undefined) while

g

sits at

-1

on the left and

+1

on the right (a jump at

x = 0

). Two different failures, two different names — Section "Non-examples" below makes the taxonomy exact.

A hole is invisible on most graphs and most calculators. Continuity is decided by

values and limits, never by how smooth the picture looks.

Layer 2 — The three-condition test

f

is continuous at $x = a$ exactly when all three hold:

$f(a)$ is defined (the point is in the domain).
$\lim_{x \to a} f(x)$ exists (left and right limits exist and agree).
They are equal: $\lim_{x \to a} f(x) = f(a)$ .

In one line:

f \text{ continuous at } a \iff \lim_{x \to a} f(x) = f(a)

with the understanding that writing the right-hand side already asserts the left side exists and

f(a)

is defined.

One-sided continuity:

f

is continuous from the right at

a

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

, from the left if

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

. Endpoints of closed intervals only need the inward side.

Continuity on an interval:

f

is continuous on

(a, b)

if it is continuous at every interior point; continuous on

[a, b]

if additionally continuous from the right at

a

and from the left at

b

Layer 3 — The continuity toolkit

Decision ladder (run top to bottom at every checkpoint question):

Is $f(a)$ defined? No → discontinuous (classify: hole, jump, or blow-up).
Does $\lim_{x \to a} f(x)$ exist? No → jump or oscillation; discontinuous.
Are they equal? No → removable discontinuity. Yes → continuous.

Combinations preserve continuity: if

f

and

g

are continuous at

a

, so are

f + g

f - g

f \cdot g

f / g

(where

g(a) \neq 0

), and

f \circ g

(where the composition is defined). This single fact is why step 3 of the ladder usually collapses to direct substitution.

Functions continuous everywhere on their domains:

Family	Why it never breaks
Polynomials	built from $x$ by addition and multiplication
Rational functions	quotient rule above, except where the denominator is $0$
$\sin x$ , $\cos x$	oscillation stays bounded and unbroken
$e^x$ , $\ln x$ ( $x > 0$ )	no jumps on their domains
$\lvert x \rvert$ , $\sqrt[n]{x}$	corners and endpoints are still continuous points

A corner (like

\lvert x \rvert

x = 0

) is continuous but not differentiable there. Continuity does not imply differentiability — but differentiability always implies continuity. Papers test exactly this one-way street.

Worked anatomy

Example A — the piecewise junction (find the parameter)

Let

f(x) = \begin{cases} 2x + 1 & x < 3 \\ kx - 2 & x \geq 3 \end{cases}

. Find

k

f

is continuous at

x = 3

Step	Move	Why it is legal
1	Left limit: $\lim_{x \to 3^-} (2x + 1) = 7$	polynomials are continuous; substitute directly
2	Value: $f(3) = 3k - 2$	at $x = 3$ the second piece applies (note $\geq$ )
3	Right limit: $\lim_{x \to 3^+} (kx - 2) = 3k - 2$	linear in $x$ for fixed $k$ ; substitute
4	Demand $3k - 2 = 7$	continuity = limit equals value (condition 3)
5	$k = 3$	algebra

Check the boundary convention before step 2: had the second piece read

x > 3

, then

f(3)

would be undefined and no

k

could make

f

continuous at

3

. Papers plant exactly this variant.

Example B — classify and repair

Classify the discontinuity of

f(x) = \dfrac{x^2 - 4}{x - 2}

x = 2

, and repair it if possible.

Step	Move	Why it is legal
1	$f(2)$ undefined ( $0/0$ form)	condition 1 fails → discontinuous
2	$\lim_{x \to 2} \dfrac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x + 2) = 4$	factor and cancel: valid for $x \neq 2$ , and limits ignore the point itself
3	Limit exists ( $4$ ), value missing → removable	the taxonomy: exists-but-unequal/missing means removable
4	Repair: redefine $f(2) = 4$	one-point redefinition; now all three conditions hold

Non-examples: the three ways continuity fails

Failure mode	Signature	Example at the point	Repairable?
Removable — limit exists, value missing or wrong	hole	$\dfrac{x^2-4}{x-2}$ at $x = 2$	yes, redefine one point
Jump — sides exist but disagree	leap	$\dfrac{\lvert x \rvert}{x}$ at $x = 0$	no
Infinite — unbounded nearby	blow-up	$\dfrac{1}{x^2}$ at $x = 0$	no

"The limit exists, so it is continuous." Wrong — conditions 1 and 3 still have to hold. The piecewise function with

f(2) = 5

but limit

3

has a perfectly good limit and is still discontinuous.

f(a)

is defined, so it is continuous." Wrong — Example:

f(x) = \dfrac{1}{x^2}

is undefined at

0

; but even where defined, a jump between pieces breaks conditions 2–3. Defined is necessary, not sufficient.

The Intermediate Value Theorem

f

is continuous on the closed interval

[a, b]

and

N

lies strictly between

f(a)

and

f(b)

, then some

c

(a, b)

has

f(c) = N

. Intuition: an unbroken curve from one height to another must cross every height in between — no teleporting.

Root-location use. Show

x^3 - x - 1 = 0

has a root between

1

and

2

: let

f(x) = x^3 - x - 1

(polynomial, hence continuous everywhere);

f(1) = -1 < 0 < 1 = f(2)

; IVT gives

c \in (1, 2)

with

f(c) = 0

IVT needs

both hypotheses: a closed interval and continuity on it. A jump discontinuity can skip

N

entirely. And the converse is false — a discontinuous function can still hit every intermediate value (continuity is sufficient, not necessary). Papers bait the converse.

Question-pattern anatomy

#	Pattern	Recognition cue	Standard attack	Planted trap
1	Junction parameter	piecewise + "find $k$ for continuity"	equate one-sided limits with the value; mind $\geq$ vs $>$	boundary belongs to the other piece, so $f(a)$ is undefined and no $k$ works
2	Removable classify + repair	rational $0/0$ at a point	factor, cancel, limit; redefine the point	option claiming "continuous" before repair
3	Domain-gap rational	denominator zero at $a$	check limit existence first; infinite vs removable	calling a blow-up "removable"
4	Direct evaluation via continuity	trig/exp/polynomial at a point	substitute (continuity justifies it)	piecewise variant where the rule changes at $a$
5	IVT root existence	"show a root lies in $(a, b)$ "	evaluate signs at endpoints, cite both hypotheses	open interval, or a discontinuity inside
6	Corner vs break	$\lvert \cdot \rvert$ , piecewise with equal limits	limits agree $\to$ continuous; differentiability is the next question	"not differentiable, hence discontinuous"
7	Composite continuity	$f(g(x))$ at a point	continuity of outer at $g(a)$ plus inner at $a$	inner limit exists but outer breaks there
8	True/false on the one-way street	"differentiable $\Rightarrow$ continuous"	true; converse false ( $\lvert x \rvert$ at $0$ )	any option asserting the converse

Attempt-first checks

Question 1

Type: MCQ | Marks: 1 | Topic: Junction parameter

Let

f(x) = \begin{cases} x + 1 & x < 2 \\ 5 & x = 2 \\ x^2 - 1 & x > 2 \end{cases}

. Which statement is true at

x = 2

MCQ

Junction parameter

3 Unit Assessment

f

is continuous at

x = 2

because

f(2)

is defined

f

is continuous at

x = 2

because the limit exists

f

is discontinuous at

x = 2

because the limit differs from the value

f

is discontinuous at

x = 2

because the limit does not exist

Question 2

Type: MCQ | Marks: 1 | Topic: Discontinuity classification

f(x) = \dfrac{\lvert x \rvert}{x}

x = 0

has which discontinuity?

MCQ

Discontinuity classification

3 Unit Assessment

Removable

Jump

Infinite

None — it is continuous

Question 3

Type: NAT | Marks: 1 | Topic: Junction parameter

Let

f(x) = \begin{cases} 3x - 1 & x < 1 \\ k & x = 1 \\ x^2 + 1 & x > 1 \end{cases}

. Find

k

making

f

continuous at

x = 1

. Enter the integer.

NAT

3 Unit Assessment

Question 4

Type: MCQ | Marks: 1 | Topic: IVT hypotheses

Which claim correctly uses the Intermediate Value Theorem?

MCQ

IVT hypotheses

3 Unit Assessment

f(x) = \dfrac{1}{x}

[-1, 1]

takes every value between

f(-1)

and

f(1)

g

is continuous on

[1, 2]

with

g(1) < 0 < g(2)

, some

c \in (1, 2)

has

g(c) = 0

A function hitting every intermediate value must be continuous

IVT applies on the open interval

(a, b)

without endpoint conditions

Recall deck (inline flashcards)

Three-condition test :: $f(a)$ defined, limit exists, limit equals $f(a)$
Removable discontinuity :: limit exists but value missing or wrong; fixed by redefining one point
Jump discontinuity :: left and right limits exist but disagree; not repairable
One-sided continuity from the right :: $\lim_{x \to a^+} f(x) = f(a)$
Continuity on $[a, b]$ :: continuous inside, right-continuous at $a$ , left-continuous at $b$
Corner point like $|x|$ at $0$ :: continuous but not differentiable
Differentiability implies :: continuity, but the converse fails
IVT needs :: continuity on a closed interval plus a strictly-between value
IVT converse :: false; hitting intermediate values does not imply continuity
Repairing $\frac{x^2-4}{x-2}$ at $x=2$ :: redefine $f(2) = 4$

Blank-page recall

Close the chapter. On a blank page, write: the three conditions, the three failure modes with one example each, the IVT statement with both hypotheses, and the junction-parameter attack order. Anything you cannot reproduce is tomorrow's first fifteen minutes.

Next action

Read Derivatives — Definition and Rules next — it reuses the limit-of-difference-quotient machinery this chapter justified. Drill the patterns above on the Quiz 3 desk (drill decks, then a timed mock). Spacing: blank-page recall tomorrow; missed checks again in three days.

Chapter provenance: authored 2026-09-12 to the layered-chapter standard (docs/roadmaps/2026-09-12-claude-content-ecosystem.md, syntax per docs/authoring/AUTHORING_COOKBOOK.md), scoped against the BSMA1001 syllabus vault and the imported 697 - Continuity.md (untouched; this file is its new sibling). Known gaps: no rendered graphs yet — numeric tables stand in until a visual earns its pixels.

What this note is really saying

Continuity

Where this chapter sits

How to read the symbols

Layer 1 — Intuition: the pen test, then the numbers

Layer 2 — The three-condition test

Layer 3 — The continuity toolkit

Worked anatomy

Example A — the piecewise junction (find the parameter)

Example B — classify and repair

Non-examples: the three ways continuity fails

The Intermediate Value Theorem

Question-pattern anatomy

Attempt-first checks

Question 1

Question 2

Question 3

Question 4

Recall deck (inline flashcards)

Blank-page recall

Next action

Active Recall Session