Day 4: Continuity, Discontinuity & IVT

Continuity at a Point (The Three Conditions)

Continuity is a stricter promise than 'the limit exists' — the function also has to actually be there, and match.

Core Theorem

A function

f(x)

is continuous at

x=c

if and only if:
1.

f(c)

is defined.
2.

\lim_{x \to c} f(x)

exists (meaning

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

).
3.

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

.

Watch the TikTok ExplanationContinuity at a Point→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Where is

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

discontinuous?

How to think about it

Continuity can only fail where the function is undefined, so check the domain first. Then decide which kind of failure it is: a removable hole (the limit still exists) or a genuine break.

Key idea: Check where the function is undefined first, then check whether the limit still exists there.

Full solution

1. Step 1: Find Where f is Undefined

The denominator is zero when

x - 2 = 0 \implies x = 2

. Condition 1 fails here.

2. Step 2: Check if the Limit Still Exists

Factor:

\lim_{x \to 2} \frac{(x-2)(x+1)}{x-2} = \lim_{x \to 2} (x+1)

.

3. Step 3: Evaluate the Simplified Limit

\lim_{x \to 2} (x+1) = 2+1 = 3

. The limit exists.

4. Step 4: Conclude

Since

f(2)

is undefined but the limit is 3,

f

has a removable discontinuity at

x=2

.

Step-by-Step SOP

1
Check Left and Right
For piecewise functions, always check $\lim_{x \to c^-}$ and $\lim_{x \to c^+}$ separately.
2
Verify the Point
Confirm $f(c)$ exists and matches the limit before declaring continuity.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

Let

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

. Find the value of

k

that makes

f(x)

continuous at

x=2

.

Need a hint?

Set the left-hand limit equal to the right-hand limit (which equals

f(2)

here) at

x=2

.

Show solution

1. Step 1: Evaluate the Right-Hand Limit

As

x \to 2^+

, use

3x-1

:

\lim_{x \to 2^+} f(x) = 3(2)-1 = 5

.

2. Step 2: Evaluate the Left-Hand Limit

As

x \to 2^-

, use

x^2+k

:

\lim_{x \to 2^-} f(x) = 2^2+k = 4+k

.

3. Step 3: Solve for k

For continuity, Left = Right:

4+k = 5 \implies k=1

.

Practice 02Hard

Find

a

and

b

so that

f(x) = \begin{cases} \frac{x^2-4}{x-2} & x < 2 \\ ax^2 - bx + 3 & 2 \le x < 3 \\ 2x - a + b & x \ge 3 \end{cases}

is continuous everywhere.

Need a hint?

You have two 'seams' to fix (at

x=2

and

x=3

) and two unknowns — set up one equation at each seam and solve the system.

Show solution

1. Step 1: Simplify the First Piece

For

x<2

:

\frac{x^2-4}{x-2} = x+2

, so

\lim_{x \to 2^-} f(x) = 2+2 = 4

.

2. Step 2: Match at x=2

The middle piece must approach 4 as

x \to 2^+

:

a(2)^2 - b(2) + 3 = 4 \implies 4a - 2b = 1

.

3. Step 3: Match at x=3

The middle and last pieces must agree at

x=3

:

a(3)^2 - b(3) + 3 = 2(3) - a + b \implies 9a - 3b + 3 = 6 - a + b \implies 10a - 4b = 3

.

4. Step 4: Solve the System

From Step 2:

b = \frac{4a-1}{2}

. Substituting into Step 3:

10a - 4\left(\frac{4a-1}{2}\right) = 3 \implies 10a - 8a + 2 = 3 \implies 2a = 1 \implies a = \frac{1}{2}

. Then

b = \frac{4(1/2)-1}{2} = \frac{1}{2}

.

Common Pitfalls

⚠
Assuming ContinuityNever assume $f(c) = \lim_{x \to c} f(x)$ unless the problem explicitly states the function is continuous — all three conditions must be checked.
⚠
Multi-Piece Functions Need One Equation Per SeamA 3-piece function has 2 boundary points — you need one matching equation at each seam, then solve them as a system (see Example 3).

Types of Discontinuity: Removable, Jump, Infinite, and Oscillating

Not all discontinuities look the same — learn to name what's actually broken: a missing point, a mismatched jump, a blow-up to infinity, or a limit that never settles down at all.

Core Theorem

Removable:

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

exists but

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

(or

f(a)

is undefined) — a 'hole'.
Jump:

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

— the two sides disagree, so the general limit doesn't exist.
Infinite:

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

or

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

— the function blows up, usually at a vertical asymptote.
Oscillating:

f(x)

fluctuates infinitely fast near

x=a

and never settles toward any single value, so neither one-sided limit exists at all.

Watch the TikTok ExplanationTypes of Discontinuity→Watch the TikTok ExplanationTypes of Discontinuity (Part 2)→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Classify the discontinuity of

f(x) = \begin{cases} x + 1 & x < 1 \\ 4 & x = 1 \\ 2x - 1 & x > 1 \end{cases}

at

x=1

.

How to think about it

Classify by comparing three numbers: the left-hand limit, the right-hand limit, and

f

at the point. Left

\ne

right is a jump; both infinite is infinite; left

=

right but

\ne f

is removable.

Key idea: Compute both one-sided limits first — if they agree, compare that shared value to $f(1)$ .

Full solution

1. Step 1: Evaluate the One-Sided Limits

\lim_{x \to 1^-} (x+1) = 2

and

\lim_{x \to 1^+} (2x-1) = 2

.

2. Step 2: Confirm the General Limit Exists

Since both sides agree,

\lim_{x \to 1} f(x) = 2

.

3. Step 3: Compare to f(1)

f(1) = 4

, but the limit is

2

, so

2 \neq 4

.

4. Step 4: Classify

The limit exists but doesn't match the function value — this is a removable discontinuity at

x=1

.

Worked Example 02Easy

If

\lim_{x \to a} f(x) = 7

and

f(a)

is undefined, what type of discontinuity is at

x=a

?

How to think about it

Key idea: The limit exists — the only thing 'broken' is that the point itself is missing.

Full solution

1. Conclusion

Because the limit exists but the function value is not there, it is a removable discontinuity (a hole).

Step-by-Step SOP

1
Compute Both One-Sided Limits
Always start here — this immediately tells you whether you're looking at a jump (sides disagree) or something else.
2
If Sides Agree, Compare to f(a)
A mismatch (or missing $f(a)$ ) at this stage means removable; a match means the function is actually continuous there.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Classify the discontinuity of the floor function

f(x) = \lfloor x \rfloor

at

x=3

.

Need a hint?

Think about what value

\lfloor x \rfloor

takes for

x

just below 3 versus just above 3.

Show solution

1. Step 1: Evaluate the Left-Hand Limit

For

x

slightly less than 3 (like 2.9),

\lfloor x \rfloor = 2

, so

\lim_{x \to 3^-} f(x) = 2

.

2. Step 2: Evaluate the Right-Hand Limit

For

x

slightly greater than or equal to 3 (like 3.0, 3.1),

\lfloor x \rfloor = 3

, so

\lim_{x \to 3^+} f(x) = 3

.

3. Step 3: Classify

Since

2 \neq 3

, the one-sided limits disagree — this is a jump discontinuity at every integer, including

x=3

.

Practice 02Medium

Classify the discontinuity of

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

at

x=2

.

Need a hint?

Think about what happens to

\frac{1}{x-2}

as

x

gets extremely close to 2 from either side.

Show solution

1. Step 1: Confirm f(2) is Undefined

The denominator

x-2=0

at

x=2

, so

f(2)

is undefined.

2. Step 2: Check the One-Sided Limits

\lim_{x \to 2^+} \frac{1}{x-2} = +\infty

and

\lim_{x \to 2^-} \frac{1}{x-2} = -\infty

.

3. Step 3: Classify

Because the function blows up without bound as

x \to 2

(this is exactly the vertical asymptote at

x=2

), this is an infinite discontinuity.

Practice 03Hard

Classify the discontinuity of

f(x) = \sin\left(\dfrac{1}{x}\right)

at

x=0

.

Need a hint?

You've seen this exact function before (Day 1) — what happened to the limit as

x \to 0

?

Show solution

1. Step 1: Confirm f(0) is Undefined

\frac{1}{x}

is undefined at

x=0

, so

f(0)

doesn't exist.

2. Step 2: Examine the Behavior Near x=0

As

x \to 0

,

\frac{1}{x} \to \pm\infty

, so

\sin(1/x)

cycles through every value in

[-1,1]

infinitely many times — it never approaches

+\infty

,

-\infty

, or any single number.

3. Step 3: Classify

Since neither one-sided limit exists (not even as

\pm\infty

) — the function just oscillates forever — this is an oscillating discontinuity, not an infinite one.

Common Pitfalls

⚠
Removable ≠ Undefined OnlyA removable discontinuity can also happen when $f(a)$ IS defined but simply doesn't match the limit (Example 1) — not just when it's missing entirely.
⚠
Jump Needs Both Sides to ExistA jump discontinuity requires both one-sided limits to exist individually — they just don't agree with each other.
⚠
The Floor Function Jumps at Every Integer $\lfloor x \rfloor$ isn't just discontinuous at $x=3$ — the exact same jump pattern repeats at every integer value of $x$ .
⚠
Infinite vs. Oscillating — Neither Has One-Sided Limits, But Only One 'Blows Up'Both types fail to have a finite one-sided limit, but infinite discontinuities head cleanly toward $+\infty$ or $-\infty$ (like $1/(x-2)$ ), while oscillating discontinuities never approach any value, finite or infinite (like $\sin(1/x)$ ).

Continuity of Familiar Functions

You rarely check all three continuity conditions from scratch every time — instead you lean on the fact that entire families of functions are already known to be continuous everywhere they're defined, and that combining continuous functions keeps them continuous. This is also exactly why every IVT example so far could just say 'it's a polynomial, so it's continuous' without further proof.

Core Theorem

Polynomial, rational, radical, trigonometric, inverse trigonometric, exponential, and logarithmic functions are all continuous on their entire domain.
If

f

and

g

are continuous at

x=a

, then so are

f \pm g

,

f \cdot g

,

kf

,

f^{\alpha}

, and

\frac{f}{g}

(provided

g(a) \neq 0

).
If

g

is continuous at

a

and

f

is continuous at

g(a)

, then the composite function

f(g(x))

is continuous at

a

.

Watch the TikTok ExplanationContinuity of Composite Functions→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Determine where

h(x) = \dfrac{x^2+\cos x}{x-1}

is continuous.

How to think about it

Polynomials,

e^x

,

\sin

, and

\cos

are continuous everywhere, and a quotient is continuous everywhere its denominator is nonzero. So the only points worth checking are where the bottom equals zero.

Key idea: Break the function into its numerator and denominator, then apply the quotient rule for continuity.

Full solution

1. Step 1: Check the Numerator

x^2

and

\cos x

are both continuous everywhere, so their sum

x^2+\cos x

is continuous everywhere too.

2. Step 2: Check the Denominator

x-1

is a polynomial, so it's continuous everywhere — but it equals zero at

x=1

.

3. Step 3: Apply the Quotient Rule

A quotient of continuous functions is continuous everywhere except where the denominator is zero.

4. Step 4: State the Result

h(x)

is continuous on

(-\infty, 1) \cup (1, \infty)

.

Step-by-Step SOP

1
Identify the Building Blocks
Break the function into the familiar pieces (polynomial, trig, radical, etc.) being added, multiplied, divided, or composed together.
2
Confirm Each Piece is Continuous on Its Own Domain
Each named family (polynomial, rational, radical, trig, inverse trig, exponential, logarithmic) is automatically continuous wherever it's defined.
3
Watch for Quotient and Composite Exceptions
Exclude any point where a denominator is zero, or where the inner function of a composite isn't continuous.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Determine where

h(x) = \left|\dfrac{1}{x-2}\right|

is continuous.

Need a hint?

This is a composite function — identify the inner and outer functions first.

Show solution

1. Step 1: Split Into Inner and Outer Functions

Let

f(x) = |x|

(outer) and

g(x) = \frac{1}{x-2}

(inner), so

h(x) = f(g(x))

.

2. Step 2: Check the Outer Function

f(x) = |x|

is continuous everywhere, for all real numbers.

3. Step 3: Check the Inner Function

g(x) = \frac{1}{x-2}

is continuous everywhere except

x=2

.

4. Step 4: Apply the Composite Rule

Since the outer function is continuous everywhere,

h

is continuous wherever the inner function is — that is, on

(-\infty, 2) \cup (2, \infty)

.

Common Pitfalls

⚠
'Continuous' Doesn't Mean 'Defined Everywhere' $\sqrt{x}$ and $\ln x$ are continuous — but only on their domain ( $[0,\infty)$ and $(0,\infty)$ respectively). 'Continuous' always means continuous on wherever the function is actually defined.
⚠
Quotients Need the Denominator CheckBefore declaring a quotient continuous everywhere, always find where the denominator is zero and exclude those points — even if the numerator is perfectly well-behaved.

Intermediate Value Theorem (IVT)

An 'existence' theorem used to prove that a continuous function must pass through a specific value, without ever having to solve for it exactly.

Core Theorem

If

f

is continuous on

[a, b]

and

k

is any number between

f(a)

and

f(b)

, then there exists at least one number

c

in

(a, b)

such that

f(c) = k

.
Corollary (Root-Finding): If

f

is continuous on

[a,b]

and

f(a)

,

f(b)

have opposite signs, then there exists at least one

c \in (a,b)

such that

f(c) = 0

. (This is just the theorem above with

k=0

— it's the special case you'll use most often.)

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Show that

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

has a zero on the interval

[0, 1]

.

How to think about it

To show an equation has a solution without solving it, find two inputs where

f

comes out with opposite signs. If

f

is continuous, it has to cross zero somewhere between them.

Key idea: Find $f(0)$ and $f(1)$ and check if 0 lies between them.

Full solution

1. Step 1: Check Continuity

f(x)

is a polynomial, so it is continuous everywhere, including on

[0,1]

.

2. Step 2: Evaluate Endpoints

f(0) = 0^3+0-1 = -1

.

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

.

3. Step 3: Apply IVT

Since

f(0) < 0 < f(1)

, by the IVT there is a

c \in (0,1)

such that

f(c) = 0

.

Worked Example 02Easy

A continuous function

g

has values

g(2) = 10

and

g(5) = 20

. Is there a value

x=c

such that

g(c) = 15

?

How to think about it

Key idea: Is 15 between 10 and 20?

Full solution

1. Analysis

Since

g

is continuous and

10 < 15 < 20

, the IVT guarantees at least one

c

in

(2,5)

where

g(c) = 15

.

Step-by-Step SOP

1
State Continuity
Confirm (and explicitly write) that the function has no holes or jumps on the interval in question.
2
Test the Endpoints
Evaluate $f(a)$ and $f(b)$ to find the range of $y$ -values guaranteed to be hit.
3
Conclude with the Standard Justification
Write: 'Since $f(x)$ is continuous on $[a,b]$ and $k$ is between $f(a)$ and $f(b)$ , by the IVT there is a $c$ in $(a,b)$ such that $f(c)=k$ .'

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Why can't the IVT be used for

f(x) = 1/x

on

[-1, 1]

to prove

f(c) = 0

for some

c

?

Need a hint?

Check the requirements of the theorem before applying it.

Show solution

1. Conclusion

f(x) = 1/x

is not continuous at

x=0

, which is inside the interval

[-1,1]

— the continuity requirement is violated, so the IVT cannot be applied.

Practice 02Medium

Show that the equation

4x^3 - 6x^2 + 3x - 2 = 0

has a solution between

x=1

and

x=2

.

Need a hint?

Define

f(x) = 4x^3-6x^2+3x-2

and check its sign at the two endpoints.

Show solution

1. Step 1: Check Continuity

f(x) = 4x^3-6x^2+3x-2

is a polynomial, so it's continuous on

[1,2]

.

2. Step 2: Evaluate the Endpoints

f(1) = 4-6+3-2 = -1

.

f(2) = 32-24+6-2 = 12

.

3. Step 3: Apply IVT

Since

f(1) = -1 < 0 < 12 = f(2)

, by the IVT there is a

c \in (1,2)

such that

f(c) = 0

— a solution to the equation exists in that interval.

Practice 03Medium

Prove that the curves

y = x^3

and

y = 3x+1

must intersect somewhere.

Need a hint?

Curves 'intersecting' means their difference equals zero somewhere — turn this into a root-existence problem.

Show solution

1. Step 1: Set Up a Difference Function

Let

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

. The curves intersect wherever

f(x) = 0

.

2. Step 2: Check Continuity

f(x)

is a polynomial, so it's continuous everywhere.

3. Step 3: Find a Sign Change

f(2) = 8-6-1 = 1 > 0

and

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

.

4. Step 4: Apply IVT

Since

f(0) < 0 < f(2)

and

f

is continuous on

[0,2]

, by the IVT there is a

c \in (0,2)

with

f(c) = 0

— meaning the two curves intersect at

x=c

.

Practice 04Hard

Show that any continuous function

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

must have a fixed point — a value

c

where

f(c)=c

. Then verify it for

f(x) = \dfrac{x^2+1}{3}

.

Need a hint?

You can't apply the IVT to

f

directly since you're not looking for a zero of

f

— instead, build a new function

g(x) = f(x) - x

and find a zero of that.

Show solution

1. Step 1: Build a Difference Function

Let

g(x) = f(x) - x

. A fixed point of

f

(where

f(c)=c

) is exactly a zero of

g

(where

g(c)=0

).

2. Step 2: Evaluate g at the Endpoints

Since

f

maps

[0,1]

into

[0,1]

:

f(0) \ge 0 \implies g(0) = f(0) - 0 \ge 0

. Also

f(1) \le 1 \implies g(1) = f(1) - 1 \le 0

.

3. Step 3: Apply the IVT to g

g

is continuous (it's a difference of continuous functions), and

g(0) \ge 0 \ge g(1)

, so by the IVT there is a

c \in [0,1]

with

g(c) = 0

— that is,

f(c) = c

. This proves the general fixed-point result.

4. Step 4: Verify Concretely for f(x) = (x²+1)/3

Solve

x = \frac{x^2+1}{3} \implies x^2 - 3x + 1 = 0 \implies x = \frac{3 \pm \sqrt{5}}{2}

. Taking the root inside

[0,1]

:

c = \frac{3-\sqrt{5}}{2} \approx 0.382

, and indeed

f(0.382) \approx 0.382

.

Common Pitfalls

⚠
Forgetting ContinuityYou MUST explicitly state that the function is continuous on the interval before invoking the IVT on the AP exam, or you will lose points.
⚠
The x vs y TrapThe IVT guarantees a 'c' value (an $x$ -value) exists, but you prove it using the 'y-values' (the endpoint outputs) — don't mix the two up in your justification.
⚠
'Curves Intersect' is a Root Problem in DisguiseWhenever you need to show two curves cross, subtract one from the other and apply the IVT to the difference — you rarely need to solve for the exact intersection point (see Example 5).
⚠
Fixed Point ≠ Root — Build g(x) = f(x) - x FirstYou can't apply the IVT to $f$ itself to find a fixed point, since $f$ isn't guaranteed to cross zero. The trick is always the same: define $g(x) = f(x) - x$ and find where that is zero (see Example 6).

Worked Examples

1. Step 1: Find Where f is Undefined

2. Step 2: Check if the Limit Still Exists

3. Step 3: Evaluate the Simplified Limit

4. Step 4: Conclude

Check Left and Right

Verify the Point

Practice — Your Turn

1. Step 1: Evaluate the Right-Hand Limit

2. Step 2: Evaluate the Left-Hand Limit

3. Step 3: Solve for k

1. Step 1: Simplify the First Piece

2. Step 2: Match at x=2

3. Step 3: Match at x=3

4. Step 4: Solve the System

Worked Examples

1. Step 1: Evaluate the One-Sided Limits

2. Step 2: Confirm the General Limit Exists

3. Step 3: Compare to f(1)

4. Step 4: Classify

1. Conclusion

Compute Both One-Sided Limits

If Sides Agree, Compare to f(a)

Practice — Your Turn

1. Step 1: Evaluate the Left-Hand Limit

2. Step 2: Evaluate the Right-Hand Limit

3. Step 3: Classify

1. Step 1: Confirm f(2) is Undefined

2. Step 2: Check the One-Sided Limits

3. Step 3: Classify

1. Step 1: Confirm f(0) is Undefined

2. Step 2: Examine the Behavior Near x=0

3. Step 3: Classify

Worked Examples

1. Step 1: Check the Numerator

2. Step 2: Check the Denominator

3. Step 3: Apply the Quotient Rule

4. Step 4: State the Result

Identify the Building Blocks

Confirm Each Piece is Continuous on Its Own Domain

Watch for Quotient and Composite Exceptions

Practice — Your Turn

1. Step 1: Split Into Inner and Outer Functions

2. Step 2: Check the Outer Function

3. Step 3: Check the Inner Function

4. Step 4: Apply the Composite Rule

Worked Examples

1. Step 1: Check Continuity

2. Step 2: Evaluate Endpoints

3. Step 3: Apply IVT

1. Analysis

State Continuity

Test the Endpoints

Conclude with the Standard Justification

Practice — Your Turn

1. Conclusion

1. Step 1: Check Continuity

2. Step 2: Evaluate the Endpoints

3. Step 3: Apply IVT

1. Step 1: Set Up a Difference Function

2. Step 2: Check Continuity

3. Step 3: Find a Sign Change

4. Step 4: Apply IVT

1. Step 1: Build a Difference Function

2. Step 2: Evaluate g at the Endpoints

3. Step 3: Apply the IVT to g

4. Step 4: Verify Concretely for f(x) = (x²+1)/3

Gary Chang