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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: f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}

At x = 1: lim⁡x→1f(x)=2\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)={xif x<1x+2if x≥1f(x) = \begin{cases} x & \text{if } x < 1 \\ x + 2 & \text{if } x \geq 1 \end{cases}

At x = 1:

  • lim⁡x→1−f(x)=1\lim_{x \to 1^-} f(x) = 1
  • lim⁡x→1+f(x)=3\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)=1x−2f(x) = \frac{1}{x - 2}

At x = 2: lim⁡x→2+f(x)=+∞\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

TypeLimit Exists?f(a) Exists?Can Fix It?Visual
RemovableYesMaybe, but wrongYes ✓Hole ○
JumpNo (sides differ)MaybeNo ✗Step ⌐⌙
InfiniteNo (infinite)NoNo ✗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)=x+2x2−4f(x) = \frac{x + 2}{x^2 - 4} at x = 2

Factor: x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)

f(x)=x+2(x−2)(x+2)=1x−2f(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)={x2if x≠35if x=3f(x) = \begin{cases} x^2 & \text{if } x \neq 3 \\ 5 & \text{if } x = 3 \end{cases}

lim⁡x→3f(x)=lim⁡x→3x2=9\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)=⌊x⌋f(x) = \lfloor x \rfloor (floor function) at x = 2

  • lim⁡x→2−⌊x⌋=1\lim_{x \to 2^-} \lfloor x \rfloor = 1 (approaching 2 from below)
  • lim⁡x→2+⌊x⌋=2\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

📚 Practice Problems

1Problem 1easy

❓ Question:

Identify the type of discontinuity in f(x)=x2−16x−4f(x) = \frac{x^2 - 16}{x - 4} at x = 4.

💡 Show Solution

Step 1: Check if discontinuous

f(4)=16−164−4=00f(4) = \frac{16 - 16}{4 - 4} = \frac{0}{0}

Undefined, so discontinuous! ✓

Step 2: Find the limit

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

lim⁡x→4f(x)=lim⁡x→4(x+4)=8\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)={x−1if x<0x2if 0≤x<25if x=24if x>2g(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→0−(x−1)=−1\lim_{x \to 0^-} (x - 1) = -1

Right limit: lim⁡x→0+x2=0\lim_{x \to 0^+} x^2 = 0

−1≠0-1 \neq 0 → Different one-sided limits!

Type: Jump discontinuity at x = 0


At x = 2:

Left limit: lim⁡x→2−x2=4\lim_{x \to 2^-} x^2 = 4

Right limit: lim⁡x→2+4=4\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

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

Explain using:

📋 AP Calculus AB — Exam Format Guide

⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
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💡 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

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Limits & Continuity

❓ Frequently Asked Questions

What is Types of Discontinuity?▾
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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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.
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.