Types of Discontinuity

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

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:

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

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!