Types of Discontinuity

Classifying the different ways a function can be discontinuous

Types of Discontinuity

Not all discontinuities are created equal! There are three main types.

1. Removable Discontinuity (Hole)

What it is: The limit exists, but either f(a) doesn't exist or f(a) ≠ limit

Visual: An open circle (hole) in the graph

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

At x = 1: $\lim_{x \to 1} f(x) = 2$ but f(1) is undefined.

Why "removable"? You could "fix" it by defining or redefining f(1) = 2!

2. Jump Discontinuity

What it is: Left and right limits exist but are different

Visual: The graph "jumps" from one height to another

Example: $f(x) = \begin{cases} x & \text{if } x < 1 \\ x + 2 & \text{if } x \geq 1 \end{cases}$

At x = 1:

$\lim_{x \to 1^-} f(x) = 1$
$\lim_{x \to 1^+} f(x) = 3$
These are different, so there's a jump!

Why not removable? No way to pick a single value for f(1) that makes both sides happy.

3. Infinite Discontinuity (Vertical Asymptote)

What it is: At least one one-sided limit is infinite

Visual: The graph shoots off to $\pm\infty$

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

At x = 2: $\lim_{x \to 2^+} f(x) = +\infty$ (infinite limit)

Why the worst? The function completely "blows up" - can't fix it at all!

Comparison Table

| Type | Limit Exists? | f(a) Exists? | Can Fix It? | Visual | |------|---------------|--------------|-------------|---------| | Removable | Yes | Maybe, but wrong | Yes ✓ | Hole ○ | | Jump | No (sides differ) | Maybe | No ✗ | Step ⌐⌙ | | Infinite | No (infinite) | No | No ✗ | Asymptote ↕ |

Identifying the Type

Step 1: Find the one-sided limits

Step 2: Apply this flowchart:

Both sides equal → Removable (if discontinuous)
Both sides different (but finite) → Jump
At least one side infinite → Infinite

Example Analysis 1

$f(x) = \frac{x + 2}{x^2 - 4}$ at x = 2

Factor: $x^2 - 4 = (x - 2)(x + 2)$

$f(x) = \frac{x + 2}{(x - 2)(x + 2)} = \frac{1}{x - 2}$ (after canceling)

At x = 2: The denominator → 0 while numerator → 1

One-sided limits are infinite!

Type: Infinite discontinuity (vertical asymptote at x = 2)

Example Analysis 2

$f(x) = \begin{cases} x^2 & \text{if } x \neq 3 \\ 5 & \text{if } x = 3 \end{cases}$

$\lim_{x \to 3} f(x) = \lim_{x \to 3} x^2 = 9$

But f(3) = 5 ≠ 9

Type: Removable discontinuity (could fix by changing f(3) to 9)

Example Analysis 3

$f(x) = \lfloor x \rfloor$ (floor function) at x = 2

$\lim_{x \to 2^-} \lfloor x \rfloor = 1$ (approaching 2 from below)
$\lim_{x \to 2^+} \lfloor x \rfloor = 2$ (at 2 and above)

Different limits!

Type: Jump discontinuity

The Intermediate Value Theorem

Important connection: If f is continuous on [a, b], it must take every value between f(a) and f(b).

Discontinuities break this! With a jump, the function "skips" values.

Practice Tip

When analyzing discontinuity:

Find where it's discontinuous (often where denominator = 0)
Calculate one-sided limits
Compare limits to each other and to f(a)
Classify the type based on your findings

📚 Practice Problems

1Problem 1easy

❓ Question:

Identify the type of discontinuity in $f(x) = \frac{x^2 - 16}{x - 4}$ at x = 4.

💡 Show Solution

Step 1: Check if discontinuous

$f(4) = \frac{16 - 16}{4 - 4} = \frac{0}{0}$

Undefined, so discontinuous! ✓

Step 2: Find the limit

Factor the numerator: $f(x) = \frac{(x - 4)(x + 4)}{x - 4} = x + 4$ (for $x \neq 4$ )

$\lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 8$

Step 3: Analyze

Limit exists and equals 8
f(4) doesn't exist
No infinite behavior
Both one-sided limits equal 8

Type: Removable discontinuity

We could "remove" the hole by defining f(4) = 8!

2Problem 2hard

❓ Question:

Classify all discontinuities of $g(x) = \begin{cases} x - 1 & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x < 2 \\ 5 & \text{if } x = 2 \\ 4 & \text{if } x > 2 \end{cases}$

💡 Show Solution

Let's check potential discontinuity points: x = 0 and x = 2

At x = 0:

Left limit: $\lim_{x \to 0^-} (x - 1) = -1$

Right limit: $\lim_{x \to 0^+} x^2 = 0$

$-1 \neq 0$ → Different one-sided limits!

Type: Jump discontinuity at x = 0

At x = 2:

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

Right limit: $\lim_{x \to 2^+} 4 = 4$

Both limits equal 4, and g(2) = 5

Limit exists (= 4) but g(2) = 5 ≠ 4

Type: Removable discontinuity at x = 2

Could fix by changing g(2) from 5 to 4!

Summary:

Jump discontinuity at x = 0
Removable discontinuity at x = 2

🎴

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