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🎯⭐ INTERACTIVE LESSON

Continuity

Learn step-by-step with interactive practice!

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

ff is continuous at x=cx = c if ALL THREE conditions hold:

  1. f(c)f(c) is defined (the point exists)
  2. lim⁡x→cf(x)\lim_{x \to c} f(x) exists (left and right limits agree)
  3. lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c) (limit equals function value)

If any condition fails → ff is discontinuous at cc.

Visual Check

ConditionWhat You See
(1) failsOpen circle — f(c)f(c) undefined
(2) failsJump — left ≠ right
(3) failsHole — limit ≠ value
All passSmooth curve through (c,f(c))(c, f(c))

Examples: Continuous or Not?

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

  1. f(3)=9f(3) = 9 ✓
  2. lim⁡x→3x2=9\lim_{x \to 3} x^2 = 9 ✓
  3. 9=99 = 9 ✓

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

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

  1. f(1)f(1) = undefined (0/00/0) ✗

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

Example 3: Piecewise

g(x)={x+1x<25x=2x+1x>2g(x) = \begin{cases} x+1 & x < 2 \\ 5 & x = 2 \\ x+1 & x > 2 \end{cases}

  1. g(2)=5g(2) = 5 ✓
  2. lim⁡x→2g(x)=3\lim_{x \to 2} g(x) = 3 ✓ (both sides approach 33)
  3. 3≠53 \neq 5 ✗

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

Continuous Functions You Know

Always Continuous (on their domains)

  • Polynomials: x2,x3+2x−1x^2, x^3 + 2x - 1, etc. — continuous everywhere
  • Rational functions: continuous where denominator ≠ 0
  • Trig functions: sin⁡x,cos⁡x\sin x, \cos x — continuous everywhere
  • Exponentials: ex,2xe^x, 2^x — continuous everywhere
  • Logarithms: ln⁡x\ln x — continuous for x>0x > 0
  • Root functions: x\sqrt{x} — continuous on domain

Key Property

Combinations of continuous functions are continuous:

  • Sum/difference: f±gf \pm g
  • Product: f⋅gf \cdot g
  • Quotient: f/gf/g (where g≠0g \neq 0)
  • Composition: f(g(x))f(g(x)) (where defined)

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

Continuity Basics Quiz 🎯

Check the three conditions for f(x)=x2−4x−2f(x) = \frac{x^2-4}{x-2} at x=2x = 2:

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

2) lim⁡x→2x2−4x−2\lim_{x \to 2}\frac{x^2-4}{x-2} = ? (simplify first):

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

Continuity Concepts 🔽

Exit Quiz ✅

Part 2: Types of Discontinuity

🔍 Types of Discontinuities

Part 2 of 7

Classification

TypeWhat HappensExample
RemovableLimit exists, but ≠ f(c)f(c) (or f(c)f(c) undefined)Hole in the graph
JumpLeft limit ≠ right limitStep function
InfiniteFunction → ±∞\pm\inftyVertical asymptote
OscillatingNo limit (wiggles)sin⁡(1/x)\sin(1/x) near 0

Removable Discontinuity

f(x)=x2−9x−3f(x) = \frac{x^2-9}{x-3} at x=3x=3

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

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

Jump Discontinuities

Definition

lim⁡x→c−f(x)≠lim⁡x→c+f(x)\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x) — the one-sided limits exist but disagree.

Classic Example: Floor Function

⌊x⌋=greatest integer≤x\lfloor x \rfloor = \text{greatest integer} \leq x

At every integer nn:

  • lim⁡x→n−⌊x⌋=n−1\lim_{x \to n^-} \lfloor x \rfloor = n - 1
  • lim⁡x→n+⌊x⌋=n\lim_{x \to n^+} \lfloor x \rfloor = n
  • Jump of size 1

Piecewise Example

h(x)={2xx<13x+1x≥1h(x) = \begin{cases} 2x & x < 1 \\ 3x + 1 & x \geq 1 \end{cases}

  • Left: lim⁡x→1−2x=2\lim_{x \to 1^-} 2x = 2
  • Right: lim⁡x→1+(3x+1)=4\lim_{x \to 1^+}(3x+1) = 4
  • 2≠42 \neq 4 → jump discontinuity of size ∣4−2∣=2|4-2| = 2

Infinite Discontinuities

Vertical Asymptotes

When f(x)→±∞f(x) \to \pm\infty as x→cx \to c, there is an infinite (essential) discontinuity.

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

  • lim⁡x→2−1x−2=−∞\lim_{x \to 2^-} \frac{1}{x-2} = -\infty
  • lim⁡x→2+1x−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)f(x) = \sin(1/x) at x=0x = 0:

  • Function oscillates between −1-1 and 11 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)=1(x−5)2f(x) = \frac{1}{(x-5)^2} at x=5x = 5:

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

3) x2−4x+2\frac{x^2-4}{x+2} at x=−2x = -2:

Discontinuity Classification 🔽

Exit Quiz ✅

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=cx = c

  1. Compute lim⁡x→c−f(x)\lim_{x \to c^-} f(x) (from the left piece)
  2. Compute lim⁡x→c+f(x)\lim_{x \to c^+} f(x) (from the right piece)
  3. Compute f(c)f(c) (which piece defines it?)
  4. Check: left limit = right limit = f(c)f(c)?

Worked Examples

Example 1: Continuous

f(x)={x2x≤24x−4x>2f(x) = \begin{cases} x^2 & x \leq 2 \\ 4x - 4 & x > 2 \end{cases}

At x=2x = 2:

  • Left: lim⁡x→2−x2=4\lim_{x \to 2^-} x^2 = 4
  • Right: lim⁡x→2+(4x−4)=4\lim_{x \to 2^+}(4x-4) = 4
  • Value: f(2)=22=4f(2) = 2^2 = 4
  • 4=4=44 = 4 = 4 ✓ Continuous!

Example 2: Not Continuous

g(x)={2x+1x<3x2x≥3g(x) = \begin{cases} 2x + 1 & x < 3 \\ x^2 & x \geq 3 \end{cases}

At x=3x = 3:

  • Left: lim⁡x→3−(2x+1)=7\lim_{x \to 3^-}(2x+1) = 7
  • Right: lim⁡x→3+x2=9\lim_{x \to 3^+} x^2 = 9
  • 7≠97 \neq 9 → Jump discontinuity

Finding Values for Continuity

The Classic Problem: "Find kk so ff is continuous"

f(x)={3x+kx<2x2+1x≥2f(x) = \begin{cases} 3x + k & x < 2 \\ x^2 + 1 & x \geq 2 \end{cases}

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

  • Left: lim⁡x→2−(3x+k)=6+k\lim_{x \to 2^-}(3x+k) = 6 + k
  • Right: lim⁡x→2+(x2+1)=5\lim_{x \to 2^+}(x^2+1) = 5
  • Set equal: 6+k=5⇒k=−16 + k = 5 \Rightarrow k = -1

Two Parameters: "Find aa and bb"

f(x)={2xx≤1ax+b1<x<35xx≥3f(x) = \begin{cases} 2x & x \leq 1 \\ ax + b & 1 < x < 3 \\ 5x & x \geq 3 \end{cases}

At x=1x = 1: 2(1)=a(1)+b⇒a+b=22(1) = a(1) + b \Rightarrow a + b = 2

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

Subtract: 2a=13⇒a=6.5,b=−4.52a = 13 \Rightarrow a = 6.5, b = -4.5

Piecewise Continuity Quiz 🎯

Find the value that makes each continuous:

1) {2x+kx<43xx≥4\begin{cases} 2x+k & x < 4 \\ 3x & x \geq 4 \end{cases}. Find kk:

2) {x2x≤1mx+bx>1\begin{cases} x^2 & x \leq 1 \\ mx + b & x > 1 \end{cases} with m=3m=3. Find bb for continuity at x=1x=1:

3) {5x<0ax2+5x≥0\begin{cases} 5 & x < 0 \\ ax^2 + 5 & x \geq 0 \end{cases}. Left limit at x=0x=0:

Piecewise Analysis 🔽

Exit Quiz ✅

Part 4: Piecewise Continuity

🎯 Continuity on Intervals

Part 4 of 7

Continuous on an Interval

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

ff is continuous on [a,b][a, b] if:

  • Continuous on (a,b)(a, b)
  • lim⁡x→a+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a) (right-continuous at left endpoint)
  • lim⁡x→b−f(x)=f(b)\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][a,b]. Endpoints must be included.

Examples

  • f(x)=xf(x) = \sqrt{x}: continuous on [0,∞)[0, \infty)
  • f(x)=ln⁡xf(x) = \ln x: continuous on (0,∞)(0, \infty)
  • f(x)=1/xf(x) = 1/x: continuous on (0,∞)(0, \infty) and (−∞,0)(-\infty, 0) but NOT on any interval containing 00

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

FunctionDomainContinuous on domain?
xnx^nR\mathbb{R}Yes
1/x1/xx≠0x \neq 0Yes
x\sqrt{x}x≥0x \geq 0Yes
ln⁡x\ln xx>0x > 0Yes
tan⁡x\tan xx≠π/2+nπx \neq \pi/2 + n\piYes

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)p(x)/q(x): Continuous everywhere except where q(x)=0q(x) = 0.
  • Composed: If ff and gg are continuous, so is f(g(x))f(g(x)) (in domain).

Theorems Requiring Continuity

Intermediate Value Theorem (IVT)

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

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

Extreme Value Theorem (EVT)

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

Requires: continuous + closed interval.

Why These Fail Without Continuity

f(x)={1x=00x≠0f(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} on [−1,1][-1,1]

  • Discontinuous at x=0x=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)=x−2f(x) = \sqrt{x-2}. Continuous for x≥x \geq ?

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

3) f(x)=1x2−9f(x) = \frac{1}{x^2-9}. Discontinuous at x=3x = 3 and x=x = ?

Interval Concepts 🔽

Exit Quiz ✅

Part 5: Continuity & Limits

📊 IVT Applications

Part 5 of 7

Intermediate Value Theorem — Full Statement

If ff is continuous on [a,b][a,b] and NN is any number strictly between f(a)f(a) and f(b)f(b), then there exists at least one c∈(a,b)c \in (a,b) such that f(c)=Nf(c) = N.

What You Can Prove with IVT

  1. Existence of roots: f(a)f(a) and f(b)f(b) have opposite signs → there's a zero in (a,b)(a,b)
  2. Existence of specific values: ff must hit every value between f(a)f(a) and f(b)f(b)
  3. 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 cc is (just somewhere in (a,b)(a,b))
  • Anything about discontinuous functions

Proving Roots Exist

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

Let f(x)=x3+x−1f(x) = x^3 + x - 1.

  • f(0)=0+0−1=−1<0f(0) = 0 + 0 - 1 = -1 < 0
  • f(1)=1+1−1=1>0f(1) = 1 + 1 - 1 = 1 > 0
  • ff is a polynomial → continuous

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

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

Let g(x)=cos⁡x−xg(x) = \cos x - x.

  • g(0)=1−0=1>0g(0) = 1 - 0 = 1 > 0
  • g(π/2)=0−π/2≈−1.57<0g(\pi/2) = 0 - \pi/2 \approx -1.57 < 0
  • gg is continuous

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

Template for IVT Proofs

  1. Define f(x)f(x) (often rearrange to f(x)=0f(x) = 0 form)
  2. State that ff is continuous on [a,b][a,b] (and why)
  3. Compute f(a)f(a) and f(b)f(b) → show they have opposite signs (or bracket target)
  4. Conclude by IVT

The Bisection Method

Finding Roots Numerically

IVT says a root exists. Bisection narrows it down:

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

StepIntervalMidpoint mmf(m)f(m)New Interval
1[1,2][1, 2]1.51.50.25>00.25 > 0[1,1.5][1, 1.5]
2[1,1.5][1, 1.5]1.251.25−0.4375<0-0.4375 < 0[1.25,1.5][1.25, 1.5]
3[1.25,1.5][1.25, 1.5]1.3751.375−0.109<0-0.109 < 0[1.375,1.5][1.375, 1.5]
4[1.375,1.5][1.375, 1.5]1.43751.43750.066>00.066 > 0[1.375,1.4375][1.375, 1.4375]

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

Each step halves the interval. After nn steps, error <b−a2n< \frac{b-a}{2^n}.

IVT Applications Quiz 🎯

IVT Practice:

1) f(x)=x2−5f(x) = x^2 - 5. f(2)=?f(2) = ?:

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

3) Since f(2)f(2) and f(3)f(3) have opposite signs, a root of x2−5=0x^2-5=0 is between 2 and 3. This root is ?\sqrt{?}:

IVT Concepts 🔽

Exit Quiz ✅

Part 6: Problem-Solving Workshop

🔬 Continuity & Limits — Deep Connections

Part 6 of 7

Continuity IS a Limit Statement

The definition of continuity at cc is exactly:

lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c)

This single equation packs all three conditions:

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

Composites and Continuity

If gg is continuous at cc and ff is continuous at g(c)g(c), then f∘gf \circ g is continuous at cc.

lim⁡x→cf(g(x))=f(lim⁡x→cg(x))=f(g(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 ff is continuous: lim⁡x→cf(g(x))=f(lim⁡x→cg(x))\lim_{x \to c} f(g(x)) = f(\lim_{x \to c} g(x))

Example 1

lim⁡x→04+sin⁡x=lim⁡x→0(4+sin⁡x)=4+0=2\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 x\sqrt{\phantom{x}} because square root is continuous.

Example 2

lim⁡x→1ex2−1=elim⁡x→1(x2−1)=e0=1\lim_{x \to 1} e^{x^2-1} = e^{\lim_{x \to 1}(x^2-1)} = e^0 = 1

We pulled the limit inside exe^{\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→2⌊x⌋≠⌊lim⁡x→2x⌋\lim_{x \to 2} \lfloor x \rfloor \neq \lfloor \lim_{x \to 2} x \rfloor when the limit is at a discontinuity of ⌊⋅⌋\lfloor \cdot \rfloor.

Special Cases

Absolute Value and Continuity

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

Continuous ≠ Differentiable

Continuity is necessary for differentiability, but NOT sufficient.

Continuous but Not Differentiable Examples

  • ∣x∣|x| at x=0x = 0 (sharp corner)
  • x3\sqrt[3]{x} at x=0x = 0 (vertical tangent)
  • xsin⁡(1/x)x\sin(1/x) at x=0x = 0 (if defined as 0 there)

Differentiable → Continuous (Always True!)

If ff is differentiable at cc, then ff is continuous at cc.

Proof sketch: f(x)−f(c)=f(x)−f(c)x−c⋅(x−c)→f′(c)⋅0=0f(x) - f(c) = \frac{f(x)-f(c)}{x-c} \cdot (x-c) \to f'(c) \cdot 0 = 0

So lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c). ✓

The Hierarchy

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

Deep Connections Quiz 🎯

Evaluate using continuity:

1) lim⁡x→πcos⁡(x)\lim_{x \to \pi} \cos(x) = ?

2) lim⁡x→4x+5\lim_{x \to 4} \sqrt{x + 5} = ?

3) lim⁡x→0e3x\lim_{x \to 0} e^{3x} = ?

Deep Connections 🔽

Exit Quiz ✅

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 $\lim_{x \to 3} (2x^2-x)$. 

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

  3) Compute $\lim_{x \to 4} \frac{x^2-16}{x-4}$.

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.

Applied mixed questions (2 questions)