Day 4

Continuity And Ivt

Continuity at a Point (The Three Conditions)Types of Discontinuity: Removable and Jump Intermediate Value Theorem (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→

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 Exercises

Example 01Medium

Where is

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

discontinuous?

Watch the TikTok ExplanationContinuity at a Point→

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Find Where f is Undefined

The denominator is zero when

x - 2 = 0 \implies x = 2

. Condition 1 fails here.

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)

Step 3: Evaluate the Simplified Limit

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

. The limit exists.

Step 4: Conclude

Since

f(2)

is undefined but the limit is 3,

f

has a **removable discontinuity** at

x=2

Example 02Easy

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 DETAILED EXPLANATION

Step 1: Evaluate the Right-Hand Limit

x \to 2^+

, use

3x-1

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

Step 2: Evaluate the Left-Hand Limit

x \to 2^-

, use

x^2+k

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

Step 3: Solve for k

For continuity, Left = Right:

4+k = 5 \implies k=1

Example 03Hard

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 DETAILED EXPLANATION

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

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

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

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 and Jump

Not all discontinuities look the same — learn to name what's actually broken: a missing point, or a mismatched jump.

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.

Watch the TikTok ExplanationTypes of Discontinuity→

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 Exercises

Example 01Medium

Classify the discontinuity of

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

x=1

Watch the TikTok ExplanationTypes of Discontinuity→

NEED A HINT?

Compute both one-sided limits first — if they agree, compare that shared value to

f(1)

SHOW DETAILED EXPLANATION

Step 1: Evaluate the One-Sided Limits

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

and

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

Step 2: Confirm the General Limit Exists

Since both sides agree,

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

Step 3: Compare to f(1)

f(1) = 4

, but the limit is

2

, so

2 \neq 4

Step 4: Classify

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

x=1

Example 02Easy

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

and

f(a)

is undefined, what type of discontinuity is at

x=a

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Conclusion

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

Example 03Medium

Classify the discontinuity of the floor function

f(x) = \lfloor x \rfloor

x=3

NEED A HINT?

Think about what value

\lfloor x \rfloor

takes for

x

just below 3 versus just above 3.

SHOW DETAILED EXPLANATION

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

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

Step 3: Classify

Since

2 \neq 3

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

x=3

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 $x$ .

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

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

(a, b)

such that

f(c) = k

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 Exercises

Example 01Easy

Show that

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

has a zero on the interval

[0, 1]

NEED A HINT?

Find

f(0)

and

f(1)

and check if 0 lies between them.

SHOW DETAILED EXPLANATION

Step 1: Check Continuity

f(x)

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

[0,1]

Step 2: Evaluate Endpoints

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

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

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

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

NEED A HINT?

Is 15 between 10 and 20?

SHOW DETAILED EXPLANATION

Analysis

Since

g

is continuous and

10 < 15 < 20

, the IVT guarantees at least one

c

(2,5)

where

g(c) = 15

Example 03Medium

Why can't the IVT be used for

f(x) = 1/x

[-1, 1]

to prove

f(c) = 0

for some

c

NEED A HINT?

Check the requirements of the theorem before applying it.

SHOW DETAILED EXPLANATION

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.

Example 04Medium

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 DETAILED EXPLANATION

Step 1: Check Continuity

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

is a polynomial, so it's continuous on

[1,2]

Step 2: Evaluate the Endpoints

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

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

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.

Example 05Medium

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 DETAILED EXPLANATION

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

Step 2: Check Continuity

f(x)

is a polynomial, so it's continuous everywhere.

Step 3: Find a Sign Change

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

and

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

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

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).

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

Discuss with Gary on Reddit

Check Left and Right

Verify the Point

Practice Exercises

Step 1: Find Where f is Undefined

Step 2: Check if the Limit Still Exists

Step 3: Evaluate the Simplified Limit

Step 4: Conclude

Step 1: Evaluate the Right-Hand Limit

Step 2: Evaluate the Left-Hand Limit

Step 3: Solve for k

Step 1: Simplify the First Piece

Step 2: Match at x=2

Step 3: Match at x=3

Step 4: Solve the System

Compute Both One-Sided Limits

If Sides Agree, Compare to f(a)

Practice Exercises

Step 1: Evaluate the One-Sided Limits

Step 2: Confirm the General Limit Exists

Step 3: Compare to f(1)

Step 4: Classify

Conclusion

Step 1: Evaluate the Left-Hand Limit

Step 2: Evaluate the Right-Hand Limit

Step 3: Classify

State Continuity

Test the Endpoints

Conclude with the Standard Justification

Practice Exercises

Step 1: Check Continuity

Step 2: Evaluate Endpoints

Step 3: Apply IVT

Analysis

Conclusion

Step 1: Check Continuity

Step 2: Evaluate the Endpoints

Step 3: Apply IVT

Step 1: Set Up a Difference Function

Step 2: Check Continuity

Step 3: Find a Sign Change

Step 4: Apply IVT

Gary Chang