🎯⭐ INTERACTIVE LESSON

Continuity

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Continuity - Complete Interactive Lesson

Part 1: Continuity Basics

🔗 What Is Continuity?

Part 1 of 7

The Intuitive Idea

A function is continuous if you can draw its graph without lifting your pen.

The Formal Definition

$f$ is continuous at $x = c$ if ALL THREE conditions hold:

$f(c)$ is defined (the point exists)
$\lim_{x \to c} f(x)$ exists (left and right limits agree)
$\lim_{x \to c} f (x)$ (limit equals function value)

If any condition fails → $f$ is discontinuous at $c$ .

Visual Check

Condition	What You See
(1) fails	Open circle — $f(c)$ undefined
(2) fails	Jump — left ≠ right
(3) fails	Hole — limit ≠ value
All pass	Smooth curve through $(c, f(c))$

Examples: Continuous or Not?

Example 1: $f(x) = x^2$ at $x = 3$

$f(3) = 9$ ✓

Continuous Functions You Know

Always Continuous (on their domains)

Polynomials: $x^2, x^3 + 2x - 1$ , etc. — continuous everywhere
Rational functions: continuous where denominator ≠ 0
Trig functions: $\sin x, \cos$ — continuous everywhere

Continuity Basics Quiz 🎯

Check the three conditions for $f(x) = \frac{x^2-4}{x-2}$ at :

Continuity Concepts 🔽

Exit Quiz ✅

Part 2: Types of Discontinuity

🔍 Types of Discontinuities

Part 2 of 7

Classification

Type	What Happens	Example
Removable	Limit exists, but ≠ $f(c)$ (or $f(c)$ undefined)	Hole in the graph
Jump	Left limit ≠ right limit	Step function
Infinite	Function →

Part 3: Intermediate Value Theorem

🧩 Piecewise Continuity

Part 3 of 7

The Key Question

For a piecewise function, continuity at the boundary is the issue. The pieces are usually nice functions — it's the join points that may fail.

Checking Continuity at a Boundary $x = c$

Compute $\lim_{x \to c^-} f(x)$ (from the left piece)

Part 4: Piecewise Continuity

🎯 Continuity on Intervals

Part 4 of 7

Continuous on an Interval

$f$ is continuous on $(a, b)$ if it is continuous at every point in $(a, b)$ .

$f$ is if:

Part 5: Continuity & Limits

📊 IVT Applications

Part 5 of 7

Intermediate Value Theorem — Full Statement

If $f$ is continuous on $[a,b]$ and $N$ is any number strictly between $f(a)$ and , then there exists at least one such that .

Part 6: Problem-Solving Workshop

🔬 Continuity & Limits — Deep Connections

Part 6 of 7

Continuity IS a Limit Statement

The definition of continuity at $c$ is exactly:

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

Part 7: Review & Applications

Continuity: Continuity synthesis across mixed function types

  **Part 7 of 7**
  
  This part focuses on solving mixed continuity exam sets. Keep notation precise and connect each symbolic step to geometric or functional meaning.
  
  ### Core definitions
  - **IVT**: continuous functions on closed intervals take all intermediate values
  - **piecewise function**: rule changes across intervals of the domain
  - **limit**: value approached by a function as input approaches a target
  
  
  ### Worked Example
  Part 7 uses direct precalculus notation to move from structure to computation.
  
  Start with a model statement, substitute known values, and simplify step by step using exact form first.
  When needed, convert to decimals only after the symbolic setup is complete.

Multiple-choice check (2 questions)

Deep-Dive: formulas and decision rules

  Use this table to pick the right expression before computing.
  
  | Tool | Formula | Best use |
  |---|---|---|
  | One-sided match | $lim_{x\to a^-}f(x)=lim_{x\to a^+}f(x)$ | two-sided existence |
  | Rational hole repair | $\frac{x^2-c^2}{x-c}=x+c;(x\neq c)$ | removable discontinuity cleanup |
  | Continuity test | $lim_{x\to a} f(x) = f(a)$ | pointwise verification |
  | Average rate | $\frac{f(b)-f(a)}{b-a}$ | bridge to local behavior |
  
  ### Common pitfalls
  - A defined value at $x=a$ does not guarantee continuity.
  - Do not classify a vertical asymptote as removable.
  - For piecewise functions, evaluate left limit, right limit, and value separately.
  
  ### Precision checks
  1. Identify givens and unknowns before selecting a formula.
  2. Keep exact values through symbolic simplification when possible.
  3. Verify units, angle mode, or domain constraints before finalizing.

Input Practice — Continuity and Limits

  1) Compute $lim_{x	o 3} (2x^2-x)$. 

  2) Compute $
  rac{f(5)-f(2)}{5-2}$ for $f(x)=x^2$.

  3) Compute $lim_{x	o 4}
  rac{x^2-16}{x-4}$.

Dropdown-select practice (3 prompts)

Strategy: graphing, calculator, and exam tactics

  **Graphing tactics**
  - Sketch anchor points or intercept behavior before detailed algebra.
  - Use symmetry, domain limits, and asymptotes to verify shape quickly.
  
  **Calculator tactics**
  - Confirm angle mode before trig operations.
  - Store intermediate values to avoid rounded drift.
  - Use table mode to test reasonableness around key inputs.
  
  **Exam tactics**
  - Translate words to symbols first, then choose the matching formula family.
  - Eliminate options that violate domain or structure.
  - If two choices are close, substitute back into the original relationship.
  
  Tie each step to IVT, piecewise function, and limit so your reasoning is explicit and checkable.

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

\lim_{x \to 3} x^2 = 9

✓

Continuous at $x = 3$ . (Polynomials are continuous everywhere!)

Example 2: $f(x) = \frac{x^2-1}{x-1}$ at $x = 1$

$f(1)$ = undefined ( $0/0$ ) ✗

Discontinuous — condition (1) fails. (But the limit exists: $\frac{(x-1)(x+1)}{x-1} \to 2$ .)

Example 3: Piecewise

$g(x) = \begin{cases} x+1 & x < 2 \\ 5 & x = 2 \\ x+1 & x > 2 \end{cases}$

$g(2) = 5$ ✓
$\lim_{x \to 2} g(x) = 3$ ✓ (both sides approach $3$ )
$3 \neq 5$ ✗

Discontinuous — removable discontinuity (hole). Could be "fixed" by redefining $g(2) = 3$ .

Exponentials:

e^x, 2^x

— continuous everywhere

Logarithms:

\ln x

— continuous for

x > 0

Root functions:

\sqrt{x}

— continuous on domain

Combinations of continuous functions are continuous:

Sum/difference: $f \pm g$
Product: $f \cdot g$
Quotient: $f/g$ (where $g \neq 0$ )
Composition: $f(g(x))$ (where defined)

This means you rarely need the three-step check for "nice" functions — just verify you're in the domain.

1) Is $f(2)$ defined? Enter "yes" or "no":

2) $\lim_{x \to 2}\frac{x^2-4}{x-2}$ = ? (simplify first):

3) Is $f$ continuous at $x=2$ ? Enter "yes" or "no":

Removable Discontinuity

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

Limit: $\frac{(x-3)(x+3)}{x-3} \to 6$ . But $f(3)$ is undefined.

"Removable" because we could define $f(3)=6$ to make it continuous.

Jump Discontinuities

Definition

$\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$ — the one-sided limits exist but disagree.

Classic Example: Floor Function

$\lfloor x \rfloor = \text{greatest integer} \leq x$

At every integer $n$ :

$\lim_{x \to n^-} \lfloor x \rfloor = n - 1$
$\lim_{x \to n^{}}$

Piecewise Example

$h(x) = \begin{cases} 2x & x < 1 \\ 3x + 1 & x \geq 1 \end{cases}$

Left: $\lim_{x \to 1^-} 2x = 2$
Right: $\lim_{x \to 1^+}(3x+1) = 4$

Infinite Discontinuities

Vertical Asymptotes

When $f(x) \to \pm\infty$ as $x \to c$ , there is an infinite (essential) discontinuity.

Example: $f(x) = \frac{1}{x-2}$

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

This is NOT removable — the function blows up.

Oscillating Discontinuity

$f(x) = \sin(1/x)$ at $x = 0$ :

Function oscillates between $-1$ and $1$ infinitely often
No limit exists (not even one-sided)
Cannot be removed

Summary: Can It Be Fixed?

Removable: YES — redefine one point
Jump: NO — would need to "teleport"
Infinite: NO — function goes to infinity
Oscillating: NO — no stable value

Discontinuity Types Quiz 🎯

Classify the discontinuity (enter: removable, jump, or infinite):

1) $f(x) = \frac{1}{(x-5)^2}$ at $x = 5$ :

2) Piecewise: left limit = 3, right limit = 7 at $x = 2$ :

3) $\frac{x^2-4}{x+2}$ at $x = -2$ :

Discontinuity Classification 🔽

Compute

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

(from the right piece)

Compute

f(c)

(which piece defines it?)

Check: left limit = right limit =

f(c)

Worked Examples

Example 1: Continuous

$f(x) = \begin{cases} x^2 & x \leq 2 \\ 4x - 4 & x > 2 \end{cases}$

At $x = 2$ :

Left: $\lim_{x \to 2^-} x^2 = 4$
Right: $\lim_{x \to^{}}$

Example 2: Not Continuous

$g(x) = \begin{cases} 2x + 1 & x < 3 \\ x^2 & x \geq 3 \end{cases}$

At $x = 3$ :

Left: $\lim_{x \to 3^-}(2x+1) = 7$
Right: $\lim_{x \to^{}}$

Finding Values for Continuity

The Classic Problem: "Find $k$ so $f$ is continuous"

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

Strategy: Set left limit = right limit at $x = 2$ :

Left: $\lim_{x \to 2^-}(3x+k) = 6 + k$

Two Parameters: "Find $a$ and $b$ "

$f(x) = \begin{cases} 2x & x \leq 1 \\ ax + b & 1 < x < 3 \\ 5x & x \geq 3 \end{cases}$

At $x = 1$ : $2(1) = a(1) + b \Rightarrow a + b = 2$

At $x = 3$ : $a(3) + b = 5(3) \Rightarrow 3a + b = 15$

Subtract: $2a = 13 \Rightarrow a = 6.5, b = -4.5$

Piecewise Continuity Quiz 🎯

Find the value that makes each continuous:

1) $\begin{cases} 2x+k & x < 4 \\ 3x & x \geq 4 \end{cases}$ . Find $k$ :

2) $\begin{cases} x^2 & x \leq 1 \\ mx + b & x > 1 \end{cases}$ with . Find for continuity at :

3) $\begin{cases} 5 & x < 0 \\ ax^2 + 5 & x \geq 0 \end{cases}$ . Left limit at :

Piecewise Analysis 🔽

continuous on $[a, b]$

Continuous on $(a, b)$
$\lim_{x \to a^+} f(x) = f(a)$ (right-continuous at left endpoint)
$\lim_{x \to b^-} f(x) = f(b)$ (left-continuous at right endpoint)

Why Closed Intervals Matter

Many theorems (IVT, EVT, MVT) require continuity on a closed interval $[a,b]$ . Endpoints must be included.

$f(x) = \sqrt{x}$ : continuous on $[0, \infty)$
$f(x) = \ln x$ : continuous on $(0, \infty)$
$f(x) = 1/x$ : continuous on $(0, \infty)$ and $(-\infty, 0)$ but NOT on any interval containing $0$

Continuity and Domain

Key Principle

A function is continuous on its domain if it has no discontinuities within that domain.

Functions Continuous on Their Entire Domain

Function	Domain	Continuous on domain?
$x^n$	$\mathbb{R}$	Yes
$1/x$	$x \neq 0$	Yes
$\sqrt{x}$	$x \geq 0$	Yes
$\ln x$	$x > 0$	Yes
$\tan x$	$x \neq \pi/2 + n\pi$	Yes

All of these are continuous where they're defined. The "discontinuities" are really just domain restrictions.

Continuity Tests for Common Types

Polynomial: Always continuous. No test needed.
Rational $p(x)/q(x)$ : Continuous everywhere except where $q(x) = 0$ .
Composed: If $f$ and are continuous, so is (in domain).

Theorems Requiring Continuity

Intermediate Value Theorem (IVT)

If $f$ is continuous on $[a,b]$ , then $f$ takes every value between $f(a)$ and $f(b)$ .

Requires: continuity on $[a,b]$ .

Extreme Value Theorem (EVT)

If $f$ is continuous on $[a,b]$ , then $f$ has an absolute maximum and absolute minimum on $[a,b]$ .

Requires: continuous + closed interval.

Why These Fail Without Continuity

$f(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases}$ on

Discontinuous at $x=0$
Never takes values between 0 and 1 (violates IVT spirit)
Still has max/min, but pathological cases can fail EVT without continuity

Interval Continuity Quiz 🎯

Determine domains of continuity:

1) $f(x) = \sqrt{x-2}$ . Continuous for $x \geq$ ?

2) $f(x) = \ln(x+5)$ . Continuous for $x >$ ?

3) $f(x) = \frac{1}{x^2-9}$ . Discontinuous at $x = 3$ and ?

Interval Concepts 🔽

What You Can Prove with IVT

Existence of roots: $f(a)$ and $f(b)$ have opposite signs → there's a zero in $(a,b)$
Existence of specific values: $f$ must hit every value between $f(a)$ and $f(b)$
Bisection method: Narrow down the location of a root

What IVT Does NOT Tell You

How many solutions exist (just "at least one")
Where exactly $c$ is (just somewhere in $(a,b)$ )
Anything about discontinuous functions

Proving Roots Exist

Example 1: $x^3 + x - 1 = 0$ has a root in $(0, 1)$

Let $f(x) = x^3 + x - 1$ .

$f(0) = 0 + 0 - 1 = -1 < 0$
$f(1) = 1 + 1 - 1 = 1 > 0$

By IVT, since $f(0) < 0 < f(1)$ , there exists $c \in (0,1)$ with $f(c) = 0$ . ✓

Example 2: $\cos x = x$ has a solution

Let $g(x) = \cos x - x$ .

$g(0) = 1 - 0 = 1 > 0$
$g(\pi/2) = 0 - \pi/2 \approx -1.57 < 0$

By IVT, $g(c) = 0$ for some $c \in (0, \pi/2)$ , meaning $\cos c = c$ .

Template for IVT Proofs

Define $f(x)$ (often rearrange to $f(x) = 0$ form)
State that $f$ is continuous on $[a,b]$ (and why)

The Bisection Method

Finding Roots Numerically

IVT says a root exists. Bisection narrows it down:

Example: $f(x) = x^2 - 2$ (finding $\sqrt{2}$ )

Step	Interval	Midpoint $m$	$f(m)$	New Interval
1	$[1, 2]$	$1.5$

After just 4 steps: $\sqrt{2} \in (1.375, 1.4375)$ . Actual: $1.4142...$

Each step halves the interval. After $n$ steps, error $< \frac{b-a}{2^n}$ .

IVT Applications Quiz 🎯

IVT Practice:

1) $f(x) = x^2 - 5$ . $f(2) = ?$ :

2) $f(3) = ?$ for the same function:

3) Since $f(2)$ and $f(3)$ have opposite signs, a root of $x^2-5=0$ is between 2 and 3. This root is :

IVT Concepts 🔽

This single equation packs all three conditions:

$f(c)$ must be defined (right side exists)
The limit must exist (left side exists)
They must be equal

Composites and Continuity

If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then $f \circ g$ is continuous at $c$ .

$\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x)\right) = f(g(c))$

You can "pass the limit inside" a continuous function!

Swapping Limits and Continuous Functions

The Rule

If $f$ is continuous: $\lim_{x \to c} f(g(x)) = f(\lim_{x \to c} g(x))$

Example 1

$\lim_{x \to 0} \sqrt{4 + \sin x} = \sqrt{\lim_{x \to 0}(4 + \sin x)} = \sqrt{4 + 0} = 2$

We pulled the limit inside $\sqrt{\phantom{x}}$ because square root is continuous.

Example 2

$\lim_{x \to 1} e^{x^2-1} = e^{\lim_{x \to 1}(x^2-1)} = e^0 = 1$

We pulled the limit inside $e^{\phantom{x}}$ because the exponential is continuous.

When You CANNOT Swap

If the outer function is NOT continuous at the limit value, this doesn't work. For example, floor function: $\lim_{x \to 2} \lfloor x \rfloor \neq \lfloor \lim_{x \to 2} x \rfloor$ when the limit is at a discontinuity of .

Special Cases

Absolute Value and Continuity

$|x|$ is continuous everywhere but NOT differentiable at $x = 0$ . This is an important distinction:

Continuous ≠ Differentiable

Continuity is necessary for differentiability, but NOT sufficient.

Continuous but Not Differentiable Examples

$|x|$ at $x = 0$ (sharp corner)
$\sqrt[3]{x}$ at $x = 0$ (vertical tangent)
$x\sin(1/x)$ at $x = 0$ (if defined as 0 there)

Differentiable → Continuous (Always True!)

If $f$ is differentiable at $c$ , then $f$ is continuous at $c$ .

Proof sketch: $f(x) - f(c) = \frac{f(x)-f(c)}{x-c} \cdot (x-c) \to f'(c) \cdot 0 = 0$

So $\lim_{x \to c} f(x) = f(c)$ . ✓

The Hierarchy

$\text{Differentiable} \subset \text{Continuous} \subset \text{Has limits}$

Deep Connections Quiz 🎯

Evaluate using continuity:

1) $\lim_{x \to \pi} \cos(x)$ = ?

2) $\lim_{x \to 4} \sqrt{x + 5}$ = ?

3) $\lim_{x \to 0} e^{3x}$ = ?

Deep Connections 🔽

Applied mixed questions (2 questions)

lim_{x \to n^{+}} ⌊ x ⌋ = n

2 \neq 4

→ jump discontinuity of size

|4-2| = 2

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

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

Value:

f(2) = 2^2 = 4

4 = 4 = 4

✓ Continuous!

lim_{x \to 3^{+}} x^{2} = 9

7 \neq 9

→ Jump discontinuity

Right:

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

Set equal:

6 + k = 5 \Rightarrow k = -1

2xax+b5xx≤11<x<3x≥3

f

is a polynomial → continuous

Compute

f(a)

and

f(b)

→ show they have opposite signs (or bracket target)

\lfloor \cdot \rfloor

Continuity

Continuity - Complete Interactive Lesson

Part 1: Continuity Basics

🔗 What Is Continuity?

The Intuitive Idea

The Formal Definition

Visual Check

Examples: Continuous or Not?

Example 1: f(x)=x2f(x) = x^2f(x)=x2 at x=3x = 3x=3

Continuous Functions You Know

Always Continuous (on their domains)

Part 2: Types of Discontinuity

🔍 Types of Discontinuities

Classification

Part 3: Intermediate Value Theorem

🧩 Piecewise Continuity

The Key Question

Checking Continuity at a Boundary x=cx = cx=c

Part 4: Piecewise Continuity

🎯 Continuity on Intervals

Continuous on an Interval

Part 5: Continuity & Limits

📊 IVT Applications

Intermediate Value Theorem — Full Statement

Part 6: Problem-Solving Workshop

🔬 Continuity & Limits — Deep Connections

Continuity IS a Limit Statement

Part 7: Review & Applications

Continuity: Continuity synthesis across mixed function types

Deep-Dive: formulas and decision rules

Strategy: graphing, calculator, and exam tactics

Example 2: f(x)=x2−1x−1f(x) = \frac{x^2-1}{x-1}f(x)=x−1x2−1​ at x=1x = 1x=1

Example 3: Piecewise

Key Property

Removable Discontinuity

Jump Discontinuities

Definition

Classic Example: Floor Function

Piecewise Example

Infinite Discontinuities

Vertical Asymptotes

Example: f(x)=1x−2f(x) = \frac{1}{x-2}f(x)=x−21​

Oscillating Discontinuity

Summary: Can It Be Fixed?

Worked Examples

Example 1: Continuous

Example 2: Not Continuous

Finding Values for Continuity

The Classic Problem: "Find kkk so fff is continuous"

Two Parameters: "Find aaa and bbb"

Why Closed Intervals Matter

Examples

Continuity and Domain

Key Principle

Functions Continuous on Their Entire Domain

Continuity Tests for Common Types

Theorems Requiring Continuity

Intermediate Value Theorem (IVT)

Extreme Value Theorem (EVT)

Why These Fail Without Continuity

What You Can Prove with IVT

What IVT Does NOT Tell You

Proving Roots Exist

Example 1: x3+x−1=0x^3 + x - 1 = 0x3+x−1=0 has a root in (0,1)(0, 1)(0,1)

Example 2: cos⁡x=x\cos x = xcosx=x has a solution

Template for IVT Proofs

The Bisection Method

Finding Roots Numerically

Example: f(x)=x2−2f(x) = x^2 - 2f(x)=x2−2 (finding 2\sqrt{2}2​)

Composites and Continuity

Swapping Limits and Continuous Functions

The Rule

Example 1

Example 2

When You CANNOT Swap

Special Cases

Absolute Value and Continuity

Continuous but Not Differentiable Examples

Differentiable → Continuous (Always True!)

The Hierarchy

Example 1: $f(x) = x^2$ at $x = 3$

Checking Continuity at a Boundary $x = c$

Example 2: $f(x) = \frac{x^2-1}{x-1}$ at $x = 1$

Example: $f(x) = \frac{1}{x-2}$

The Classic Problem: "Find $k$ so $f$ is continuous"

Two Parameters: "Find $a$ and $b$ "

Example 1: $x^3 + x - 1 = 0$ has a root in $(0, 1)$

Example 2: $\cos x = x$ has a solution

Example: $f(x) = x^2 - 2$ (finding $\sqrt{2}$ )