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)=x−1
📚 Practice Problems
1Problem 1easy
❓ Question:
Identify the type of discontinuity in f(x)=x−4 at x = 4.
Explain using:
📋 AP Calculus AB — Exam Format Guide
⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
Section
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Questions
Time
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MCQ
30
60 min
33.3%
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15
45 min
16.7%
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2
30 min
16.7%
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4
60 min
33.3%
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💡 Key Test-Day Tips
✓Show all work on FRQs
✓Use proper notation
✓Check units
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⚠️ 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
Problem:
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?
Classifying the different ways a function can be discontinuous
How can I study Types of Discontinuity effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Types of Discontinuity?▾
Types of Discontinuity is part of the AP Calculus AB course on Study Mondo, specifically in the Limits & Continuity section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Types of Discontinuity?
x2−1
At x = 1: limx→1f(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)={xx+2if x<1if x≥1
At x = 1:
limx→1−f(x)=1
limx→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 ±∞
Example:f(x)=x−21
At x = 2: limx→2+f(x)=+∞ (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)=x2−4x+2 at x = 2
Factor: x2−4=(x−2)(x+2)
f(x)=(x−2)(x+2)x+2=x−21 (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)={x25if x=3if x=3
limx→3f(x)=limx→3x2=9
But f(3) = 5 ≠ 9
Type: Removable discontinuity (could fix by changing f(3) to 9)
Example Analysis 3
f(x)=⌊x⌋ (floor function) at x = 2
limx→2−⌊x⌋=1 (approaching 2 from below)
limx→2+⌊x⌋=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
x2−16
💡 Show Solution
Step 1: Check if discontinuous
f(4)=4−416−16=00
Undefined, so discontinuous! ✓
Step 2: Find the limit
Factor the numerator:
f(x)=x−4(x−4)(x+4) (for )
limx→4f(x)=limx→4(
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)=⎩⎨⎧x−1x254
💡 Show Solution
Let's check potential discontinuity points: x = 0 and x = 2
At x = 0:
Left limit: limx→0−(x−1)
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 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
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.