Types of Discontinuity
Classifying the different ways a function can be discontinuous
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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:
At x = 1: 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:
At x = 1:
- 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
Example:
At x = 2: (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
at x = 2
Factor:
(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
But f(3) = 5 ≠ 9
Type: Removable discontinuity (could fix by changing f(3) to 9)
Example Analysis 3
(floor function) at x = 2
- (approaching 2 from below)
- (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 at x = 4.
💡 Show Solution
Step 1: Check if discontinuous
Undefined, so discontinuous! ✓
Step 2: Find the limit
Factor the numerator: (for )
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
💡 Show Solution
Let's check potential discontinuity points: x = 0 and x = 2
At x = 0:
Left limit:
Right limit:
→ Different one-sided limits!
Type: Jump discontinuity at x = 0
At x = 2:
Left limit:
Right limit:
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
3Problem 3easy
❓ Question:
Identify the type of discontinuity at x = 3 for f(x) = (x² - 9)/(x - 3).
💡 Show Solution
Step 1: Check if f(3) is defined: f(3) = (9 - 9)/(3 - 3) = 0/0 (undefined)
Step 2: Find lim(x→3) f(x): Factor: (x² - 9)/(x - 3) = [(x - 3)(x + 3)]/(x - 3) = x + 3 lim(x→3) (x + 3) = 6
Step 3: Classification: • f(3) is undefined • But lim(x→3) f(x) = 6 exists • This is a hole in the graph
Step 4: Type of discontinuity: This is a REMOVABLE discontinuity (also called a point discontinuity or hole) It can be "removed" by redefining f(3) = 6
Answer: Removable discontinuity at x = 3
4Problem 4medium
❓ Question:
Classify the discontinuity at x = 2 for f(x) = { x + 1, if x < 2; 5, if x ≥ 2 }.
💡 Show Solution
Step 1: Find f(2): f(2) = 5 (defined)
Step 2: Find lim(x→2⁻): For x < 2, f(x) = x + 1 lim(x→2⁻) (x + 1) = 3
Step 3: Find lim(x→2⁺): For x ≥ 2, f(x) = 5 lim(x→2⁺) 5 = 5
Step 4: Compare: lim(x→2⁻) = 3 ≠ 5 = lim(x→2⁺) Two-sided limit doesn't exist
Step 5: Classification: • Left and right limits are different • This creates a "jump" in the graph • JUMP discontinuity
Answer: Jump discontinuity at x = 2
5Problem 5hard
❓ Question:
Determine the type of discontinuity at x = 0 for f(x) = 1/x².
💡 Show Solution
Step 1: Check if f(0) is defined: f(0) = 1/0² = 1/0 (undefined)
Step 2: Find lim(x→0) f(x): As x→0, x²→0⁺ (always positive) 1/(small positive) = large positive lim(x→0) 1/x² = +∞
Step 3: Verify from both sides: lim(x→0⁻): x² is positive, 1/x² → +∞ lim(x→0⁺): x² is positive, 1/x² → +∞ Both one-sided limits are +∞
Step 4: Classification: • Function undefined at x = 0 • Limit is infinite (not a finite number) • Vertical asymptote at x = 0 • INFINITE discontinuity
Step 5: Note: This cannot be removed by redefining f(0) because the limit doesn't exist as a finite value
Answer: Infinite discontinuity (vertical asymptote) at x = 0
📋 AP Calculus AB — Exam Format Guide
| Section | Format | Questions | Time | Weight | Calculator |
|---|---|---|---|---|---|
| Multiple Choice (No Calculator) | MCQ | 30 | 60 min | 33.3% | 🚫 |
| Multiple Choice (Calculator) | MCQ | 15 | 45 min | 16.7% | ✅ |
| Free Response (Calculator) | FRQ | 2 | 30 min | 16.7% | ✅ |
| Free Response (No Calculator) | FRQ | 4 | 60 min | 33.3% | 🚫 |
📊 Scoring: 1-5
💡 Key Test-Day Tips
- ✓Show all work on FRQs
- ✓Use proper notation
- ✓Check units
- ✓Manage your time
⚠️ Common Mistakes: Types of Discontinuity
Avoid these 4 frequent errors
🌍 Real-World Applications: Types of Discontinuity
See how this math is used in the real world
📝 Worked Example: Related Rates — Expanding Circle
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of cm/s. How fast is the area of the circle increasing when the radius is cm?
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