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Unit 1.13 – Removing Discontinuities

AP® Calculus AB & BC | Formula Reference Sheet

Core Concept: Sometimes a function has a "hole"—a single missing point—but the graph wants to be continuous there. A removable discontinuity is one you can "fix" by simply redefining the function's value at that point to equal the limit. Think of it as filling in the hole to make the graph smooth. This unit teaches you how to identify these fixable discontinuities and patch them up algebraically!

🎯 What is a Removable Discontinuity?

REMOVABLE DISCONTINUITY (HOLE)

A function \(f(x)\) has a removable discontinuity at \(x = a\) if:

  1. The limit \(\lim_{x \to a} f(x)\) exists (left and right limits are equal and finite)
  2. EITHER \(f(a)\) is undefined, OR \(f(a)\) is defined but \(f(a) \neq \lim_{x \to a} f(x)\)

In other words: The function "wants" to have a value at \(x = a\) (the limit exists), but something prevents it (undefined or wrong value). You can "remove" the discontinuity by redefining \(f(a)\) to equal the limit!

📝 Key Insight: Removable means "fixable"—you can fill the hole. Non-removable discontinuities (jumps and vertical asymptotes) cannot be fixed by redefining a single point because the limit either doesn't exist or is infinite.

Graphical Appearance

On a graph, a removable discontinuity appears as:

  • An open circle (hole) at the point \((a, L)\) where \(L = \lim_{x \to a} f(x)\)
  • The curve approaches the hole from both sides at the same height
  • Sometimes there's a filled dot at a different y-value (if \(f(a)\) is defined but wrong)
  • No vertical asymptote, no jump—just a single missing or misplaced point

❓ Why Do Removable Discontinuities Occur?

Common Causes:

1. Rational Functions with Common Factors:

When numerator and denominator share a factor \((x - a)\), direct substitution gives \(\frac{0}{0}\) (indeterminate form). After canceling the common factor, the function simplifies to something defined at \(x = a\)—but the original function was undefined there, creating a hole.

2. Piecewise Functions with Boundary Issues:

When the left and right pieces approach the same value at a boundary, but the function is defined as something different at that point (or not defined at all).

3. Deliberately Undefined Points:

Sometimes functions are defined "for \(x \neq a\)" even though the limit exists at \(a\).

🔧 The Three-Step Process to Remove a Discontinuity

STEP-BY-STEP: Removing a Removable Discontinuity

  1. STEP 1: Verify the discontinuity is removable
    • Check that \(\lim_{x \to a} f(x)\) exists (one-sided limits equal)
    • Verify that \(f(a)\) is either undefined OR \(f(a) \neq \lim_{x \to a} f(x)\)
    • If limit doesn't exist or is infinite → NOT removable!
  2. STEP 2: Find the limit value
    • For rational functions: Factor and cancel, then evaluate simplified expression at \(x = a\)
    • For piecewise: Calculate one-sided limits from each piece
    • Let \(L = \lim_{x \to a} f(x)\)
  3. STEP 3: Redefine the function
    • Create a new function \(g(x)\) that equals \(f(x)\) everywhere except at \(x = a\)
    • Set \(g(a) = L\) (the limit value)
    • Now \(g(x)\) is continuous at \(x = a\)! ✓

📐 Type 1: Removing Discontinuities in Rational Functions

Standard Process for Rational Functions

When: \(f(x) = \frac{P(x)}{Q(x)}\) and direct substitution gives \(\frac{0}{0}\)

Process:

  1. Factor both numerator and denominator completely
  2. Cancel the common factor \((x - a)\)
  3. Evaluate the simplified expression at \(x = a\) to get \(L\)
  4. Redefine: Set \(f(a) = L\) to remove the hole

Example 1: Classic Removable Discontinuity

Problem: Remove the discontinuity from \(f(x) = \frac{x^2 - 4}{x - 2}\)

Solution:

  1. Identify the problem: At \(x = 2\), we get \(\frac{0}{0}\) (undefined)
  2. Factor the numerator:
    \[ x^2 - 4 = (x - 2)(x + 2) \]
  3. Rewrite:
    \[ f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad (x \neq 2) \]
  4. Find the limit:
    \[ \lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4 \]
  5. Redefine: Create new function
    \[ g(x) = \begin{cases} \frac{x^2-4}{x-2} & x \neq 2 \\ 4 & x = 2 \end{cases} \]

Result: \(g(x)\) is now continuous at \(x = 2\) (hole removed)! ✓

Example 2: Higher-Degree Polynomial

Problem: Remove the discontinuity from \(h(x) = \frac{x^2 - 4x + 4}{x - 2}\)

Solution:

  1. At \(x = 2\): \(\frac{0}{0}\) (undefined)
  2. Factor numerator:
    \[ x^2 - 4x + 4 = (x - 2)^2 \]
  3. Simplify:
    \[ h(x) = \frac{(x-2)^2}{x-2} = x - 2 \quad (x \neq 2) \]
  4. Find limit:
    \[ \lim_{x \to 2} (x - 2) = 0 \]
  5. Patched function:
    \[ H(x) = \begin{cases} \frac{x^2-4x+4}{x-2} & x \neq 2 \\ 0 & x = 2 \end{cases} \]

Result: Continuous everywhere! ✓

Example 3: Multiple Discontinuities

Problem: Find all discontinuities of \(f(x) = \frac{x^2 - 9}{x^2 - 5x + 6}\) and remove any that are removable

Solution:

  1. Factor numerator: \(x^2 - 9 = (x - 3)(x + 3)\)
  2. Factor denominator: \(x^2 - 5x + 6 = (x - 2)(x - 3)\)
  3. Rewrite:
    \[ f(x) = \frac{(x-3)(x+3)}{(x-2)(x-3)} = \frac{x+3}{x-2} \quad (x \neq 3) \]
  4. Discontinuity at \(x = 3\): Common factor cancels → Removable
    • \(\lim_{x \to 3} f(x) = \frac{6}{1} = 6\)
    • Redefine \(f(3) = 6\) to remove hole
  5. Discontinuity at \(x = 2\): No common factor → Vertical asymptote (NOT removable)
    • Limit is infinite
    • Cannot be fixed

Conclusion: Only the hole at \(x = 3\) is removable

🧩 Type 2: Removing Discontinuities in Piecewise Functions

Piecewise Function Strategy

Goal: Make the function continuous at boundary points

For continuity at \(x = a\):

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \]

Process:

  1. Evaluate left-hand limit using the left piece
  2. Evaluate right-hand limit using the right piece
  3. If they're equal, set \(f(a)\) equal to that common value
  4. If they're different → Jump discontinuity (NOT removable!)

Example 4: Finding the Parameter Value

Problem: Find the value of \(k\) that makes \(f(x) = \begin{cases} 3x + 2 & x < 1 \\ k & x = 1 \\ x^2 + 1 & x > 1 \end{cases}\) continuous at \(x = 1\)

Solution:

  1. Left-hand limit:
    \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x + 2) = 3(1) + 2 = 5 \]
  2. Right-hand limit:
    \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 1) = 1 + 1 = 2 \]
  3. Problem: \(5 \neq 2\) — one-sided limits are different!
  4. Conclusion: NO VALUE of \(k\) will make this continuous at \(x = 1\)
  5. This is a jump discontinuity (non-removable)

Example 5: Successful Parameter Find

Problem: Find \(k\) so that \(g(x) = \begin{cases} kx + 1 & x < 4 \\ x^2 - 7 & x \geq 4 \end{cases}\) is continuous at \(x = 4\)

Solution:

  1. Left-hand limit:
    \[ \lim_{x \to 4^-} g(x) = \lim_{x \to 4^-} (kx + 1) = 4k + 1 \]
  2. Right-hand limit & function value:
    \[ \lim_{x \to 4^+} g(x) = g(4) = 4^2 - 7 = 9 \]
  3. For continuity: Set left = right
    \[ 4k + 1 = 9 \]
  4. Solve:
    \[ 4k = 8 \quad \Rightarrow \quad k = 2 \]

Answer: \(k = 2\) makes the function continuous at \(x = 4\) ✓

Example 6: Rational in Piecewise

Problem: Find \(a\) so that \(f(x) = \begin{cases} \frac{x^2-9}{x-3} & x \neq 3 \\ a & x = 3 \end{cases}\) is continuous

Solution:

  1. Factor and simplify:
    \[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x + 3 \quad (x \neq 3) \]
  2. Find limit:
    \[ \lim_{x \to 3} f(x) = \lim_{x \to 3} (x+3) = 6 \]
  3. For continuity: Set \(f(3) = \lim_{x \to 3} f(x)\)
  4. Answer: \(a = 6\)

⚖️ Removable vs. Non-Removable Discontinuities

Feature Removable (Hole) Non-Removable (Jump/Asymptote)
Limit Exists? ✓ YES (finite) ✗ NO (or infinite)
One-sided limits Equal to each other Different OR infinite
Function value Undefined OR wrong May or may not be defined
Graphical appearance Open circle (hole) Jump (gap) or vertical asymptote
Can you fix it? ✓ YES — redefine at one point ✗ NO — limit doesn't exist
How to remove Set \(f(a) = \lim_{x \to a} f(x)\) Cannot be removed
Example \(\frac{x^2-1}{x-1}\) at \(x=1\) \(\frac{1}{x}\) at \(x=0\) (asymptote)

🚫 When You CANNOT Remove a Discontinuity

A discontinuity is NOT removable if:

  1. One-sided limits are different (Jump):
    • \(\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)\)
    • The two-sided limit does not exist
    • No single value can satisfy both sides
    • Example: Step functions at integer boundaries
  2. At least one one-sided limit is infinite (Vertical Asymptote):
    • \(\lim_{x \to a^-} f(x) = \pm\infty\) or \(\lim_{x \to a^+} f(x) = \pm\infty\)
    • Function "blows up" near the point
    • Cannot redefine to a finite value
    • Example: \(\frac{1}{x-2}\) at \(x = 2\)
  3. Oscillating behavior with no limit:
    • Function oscillates infinitely fast as \(x \to a\)
    • Example: \(\sin\left(\frac{1}{x}\right)\) as \(x \to 0\)

💡 Tips, Tricks & Strategies

✅ Essential Tips

  • Always check if limit exists first: If \(\lim_{x \to a} f(x)\) doesn't exist, discontinuity is NOT removable
  • Factor completely before canceling: Show all factoring steps—AP® graders look for this!
  • State what you're doing: "Canceling common factor" or "Setting limit equal to function value"
  • Check one-sided limits for piecewise: They must be equal for discontinuity to be removable
  • Don't forget the domain: Write "for \(x \neq a\)" when simplifying
  • Verify your answer: Check that the new function is actually continuous (all three conditions)

🎯 The "Factor-Cancel-Evaluate" Method

For rational functions with \(\frac{0}{0}\) at \(x = a\):

  1. FACTOR: Factor numerator and denominator completely
  2. CANCEL: Cancel the common factor \((x - a)\)
  3. EVALUATE: Plug \(x = a\) into the simplified expression
  4. REDEFINE: That's your value to remove the hole!

Memory aid: Think "FCE" — Factor, Cancel, Evaluate!

🔥 Quick Recognition Patterns

You likely have a removable discontinuity if you see:

  • \(\frac{0}{0}\) after substitution — Strong indicator of removable hole
  • Common factor in numerator and denominator — That factor creates the hole
  • Graph shows a smooth curve with one open circle — Visual confirmation
  • Piecewise with matching one-sided limits but wrong/missing value

You likely have a NON-removable discontinuity if:

  • \(\frac{\text{nonzero}}{0}\) after substitution — Vertical asymptote
  • After canceling, denominator still = 0 — Still has asymptote
  • Piecewise with different one-sided limits — Jump discontinuity

❌ Common Mistakes to Avoid

  • Mistake 1: Assuming all discontinuities are removable—check if limit exists first!
  • Mistake 2: Forgetting to verify one-sided limits are equal before declaring "removable"
  • Mistake 3: Canceling without showing factoring—AP® wants to see your work
  • Mistake 4: Not writing "for \(x \neq a\)" when simplifying rational functions
  • Mistake 5: Setting function value = limit without first checking if limit exists
  • Mistake 6: Trying to remove a jump or asymptote (only holes are removable!)
  • Mistake 7: Plugging into original (undefined) expression instead of simplified form
  • Mistake 8: Not checking both pieces of piecewise functions at boundaries

📝 Practice Problems

For each function, determine if the discontinuity is removable. If so, remove it:

  1. \(f(x) = \frac{x^2 - 16}{x - 4}\) at \(x = 4\)
  2. \(g(x) = \frac{x + 3}{x - 1}\) at \(x = 1\)
  3. \(h(x) = \begin{cases} x^2 & x < 3 \\ k & x = 3 \\ 6x - 9 & x > 3 \end{cases}\) — Find \(k\) for continuity
  4. \(p(x) = \frac{x^3 - 8}{x - 2}\) at \(x = 2\)

Answers:

  1. Removable: Factor to \((x-4)(x+4)/(x-4) = x+4\); redefine \(f(4) = 8\)
  2. NOT removable: Limit is \(\pm\infty\) (vertical asymptote)
  3. \(k = 9\): Both one-sided limits = 9
  4. Removable: Factor \(x^3-8=(x-2)(x^2+2x+4)\); redefine \(p(2) = 12\)

✏️ AP® Exam Success Tips

What the AP® Exam Expects:

  • Show factoring explicitly: Write out complete factorization steps
  • State one-sided limits: For piecewise, calculate left and right separately
  • Justify removability: Explain why the limit exists (or doesn't)
  • Write the patched function clearly: Use proper piecewise notation
  • Never divide by zero: When simplifying, note restrictions
  • Label holes on sketches: Open circles for original, filled for patched
  • Verify continuity conditions: Check all three after redefining

Common FRQ Formats:

  1. "Find the value of k that makes f continuous at x = a"
  2. "Remove any removable discontinuities from the function"
  3. "Determine whether the discontinuity at x = a is removable. Justify."
  4. "Redefine the function so it is continuous at x = a"
  5. "For what values is the function continuous? Explain."

⚡ Quick Reference Card

Step Action Check
1. Is it removable? Check if \(\lim_{x \to a} f(x)\) exists One-sided limits equal & finite?
2. Find the limit Factor & cancel OR evaluate pieces Get value \(L\)
3. Redefine Set \(f(a) = L\) Now continuous at \(a\)?

🔗 Why This Unit Matters

Unit 1.13 connects to:

  • Unit 1.10: Identifying types of discontinuities (removable vs. non-removable)
  • Unit 1.11-1.12: Using continuity definition to verify fixes worked
  • Unit 1.14: Intermediate Value Theorem requires continuous functions
  • Unit 2: Derivatives require continuity (can't differentiate at discontinuities)
  • Unit 3: Optimization problems need continuous functions
  • Real-world applications: Modeling situations where data has gaps that need filling

Remember: A removable discontinuity is like a pothole in a smooth road—you can fill it in! The key is that the limit must exist at that point. If one-sided limits are equal, you can redefine the function value to equal the limit and "patch the hole." Use the Factor-Cancel-Evaluate method for rational functions and check one-sided limits for piecewise functions. Master this skill, and you'll turn broken functions into smooth, continuous ones! 🔧✨