Neural Sync Active

Continuity

Registry Synced

Continuity

1010 words

5 min read

Reading compass

Now · 🎯 Learning Objectives

Continuity

🎯 Learning Objectives

By the end of this topic, you will be able to:

Determine if a function is continuous at a point
Classify discontinuities as removable, jump, or infinite
Apply the Intermediate Value Theorem to find roots
Use properties of continuous functions in problem-solving

📋 Prerequisites

Limits of Functions — two-sided limits, limit laws

📖 Core Content

12.1 Intuition: Drawing Without Lifting the Pen

A function is continuous if you can draw its graph without lifting your pen from the paper. No holes, jumps, or breaks. Formally, continuity at

x = a

requires three things:

$f(a)$ is defined
$\lim_{x \to a} f(x)$ exists
$\lim_{x \to a} f(x) = f(a)$

💡 Why this matters: Continuous functions have nice properties: they attain extreme values on closed intervals, they can be integrated, and the Intermediate Value Theorem guarantees roots exist.

12.2 Definition

f

is continuous at

x = a

iff:

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

Left-continuous at $a$ :

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

Right-continuous at $a$ :

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

Continuous on an interval: Continuous at every point in the interval.

12.3 Types of Discontinuity

(Diagram)

Type	Description	Graph	Example
Removable	$\lim_{x\to a} f(x)$ exists but $\neq f(a)$ (or $f(a)$ undefined)	Hole in graph	$f(x) = \frac{x^2-4}{x-2}$ at $x=2$
Jump	$\lim_{x\to a^-} f(x) \neq \lim_{x\to a^+} f(x)$	Break/gap	$f(x) = \lfloor x \rfloor$
Infinite	$\lim_{x\to a} f(x) = \pm\infty$	Vertical asymptote	$f(x) = \frac{1}{x}$ at $x=0$

12.4 Properties

Polynomials, rational functions (where defined), $\sin x$ , $\cos x$ , $e^x$ , $\ln x$ (where defined) are all continuous on their domains.
If $f$ and $g$ are continuous at $a$ , then $f+g$ , $f-g$ , $f \cdot g$ , $f/g$ (if $g(a) \neq 0$ ) are continuous at $a$ .
If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $f \circ g$ is continuous at $a$ .

12.5 Intermediate Value Theorem (IVT)

f

is continuous on

[a, b]

and

L

is any number between

f(a)

and

f(b)

, then there exists

c \in (a, b)

such that

f(c) = L

. Key consequence: If

f(a)

and

f(b)

have opposite signs, there is at least one root in

(a, b)

Worked Example — IVT

Show that

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

has a root in

(1, 2)

. Step 1:

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

Step 2:

f(2) = 8 - 2 - 1 = 5 > 0

Step 3: Since

f

is continuous (polynomial) and

f(1) < 0 < f(2)

, by IVT there exists

c \in (1,2)

with

f(c) = 0

\boxed{\text{Root exists in } (1,2)}

📐 Key Formulas — Summary Table

Concept	Formula
Continuity at $a$	$\lim_{x \to a} f(x) = f(a)$
IVT	If $f$ continuous on $[a,b]$ , then $\forall L$ between $f(a), f(b)$ , $\exists c: f(c)=L$
Root existence	If $f(a) \cdot f(b) < 0$ , $\exists c \in (a,b)$ with $f(c) = 0$
Discontinuity types	Removable (hole), Jump (gap), Infinite (asymptote)

⚠️ Common Pitfalls

Pitfall 1: Thinking "Continuous" Means "Differentiable"

f(x) = |x|

is continuous everywhere but not differentiable at

x = 0

(sharp corner).

Pitfall 2: IVT Requires Continuity

Can't apply IVT to a discontinuous function on

[a,b]

, even if

f(a)

and

f(b)

have opposite signs.

Pitfall 3: Confusing Removable Discontinuity with "Hole"

A removable discontinuity has a limit that exists but doesn't equal the function value. It can be "removed" by redefining

f(a)

= limit value.

📝 Practice Questions

Q1: Is $f(x) = x^2 + 3x - 1$ continuous at $x=2$ ?

Polynomials are continuous everywhere. $f(2) = 4 + 6 - 1 = 9$ , $\lim_{x\to 2} f(x) = 9$ . ✓

$\boxed{\text{Yes}}$ Q2: Find the discontinuity of $f(x) = \frac{x+1}{x^2 - 1}$ .

Denominator $= 0$ at $x = \pm 1$ . Factor: $\frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$ (for $x \neq -1$ ). $x = 1$ : infinite discontinuity (vertical asymptote). $x = -1$ : removable discontinuity (cancel factor).

$\boxed{x=1 \text{ (infinite)}, x=-1 \text{ (removable)}}$ Q3: Show that $f(x) = x^3 - 2x - 5$ has a root in $(2,3)$ .

$f(2) = 8 - 4 - 5 = -1 < 0$ , $f(3) = 27 - 6 - 5 = 16 > 0$ . Since $f$ is continuous and signs differ, IVT guarantees a root.

$\boxed{\text{Root exists}}$ Q4: Classify the discontinuity of $f(x) = \frac{|x|}{x}$ at $x=0$ .

$\lim_{x\to 0^-} \frac{|x|}{x} = -1$ , $\lim_{x\to 0^+} \frac{|x|}{x} = 1$ . LHL ≠ RHL.

$\boxed{\text{Jump discontinuity}}$ **Q5: Find $k$ so that

> f(x) = \begin{cases} 2x+1 & x2 \end{cases} >

is continuous.**

$\lim_{x\to 2^-} f(x) = 2(2)+1 = 5$ , $\lim_{x\to 2^+} f(x) = 3(2)-1 = 5$ . For continuity, $f(2) = k = 5$ .

$\boxed{k = 5}$ Q6: Which elementary functions are continuous on their domains?

Polynomials, rational, trig (sin, cos), exponential, logarithmic — all continuous on their domains.

$\boxed{\text{All elementary functions}}$ Q7: Does $f(x) = \frac{1}{x^2}$ have a removable discontinuity at $x=0$ ?

$\lim_{x\to 0} \frac{1}{x^2} = \infty$ (infinite limit) → not removable.

$\boxed{\text{No — infinite discontinuity}}$ Q8: Use IVT to show $\cos x = x$ has a solution in $(0, 1)$ .

$f(x) = \cos x - x$ . $f(0) = 1 > 0$ , $f(1) \approx 0.54 - 1 = -0.46 < 0$ . Since $f$ is continuous, IVT guarantees $c \in (0,1)$ where $f(c)=0$ , i.e., $\cos c = c$ .

$\boxed{\text{Solution exists}}$

🔗 Cross-References

Previous topics: Limits of Functions, Sequences
Next topic: Derivatives
Across courses: BSMA1003 Maths 2 (multivariable continuity) Join Discord Previous7.2 Limits of FunctionsNext8.1 Derivatives — Definition & Rules