What is Continuity?
Understanding when a function is continuous at a point
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Understanding Continuity
A function is continuous if you can draw it without lifting your pencil. Mathematically, it's more precise!
The Informal Definition
A function is continuous at x = a if:
The graph has no breaks, jumps, or holes at that point
The Formal Definition
A function f is continuous at x = a if all three conditions hold:
- f(a) exists - The function is defined at a
- exists - The limit exists at a
- - The limit equals the function value
All three must be true! If even one fails, the function is discontinuous at that point.
Testing for Continuity
Example 1: Is continuous at x = 3?
Check 1: Does f(3) exist? ✓ Yes
Check 2: Does exist? ✓ Yes
Check 3: Does ? ✓ Yes
Conclusion: f is continuous at x = 3!
Example 2: Not Continuous
Is this continuous at x = 2?
Check 1: Does f(2) exist? ✓ Yes
Check 2: Does exist? ✓ Yes
Check 3: Does ? ✗ NO!
Conclusion: f is NOT continuous at x = 2 (even though f(2) exists and the limit exists!)
Continuous Everywhere
Many familiar functions are continuous everywhere:
- Polynomials: , , etc.
- Exponential functions: ,
- Sine and cosine: ,
- Square root (on its domain): for
Common Discontinuities
Functions are typically discontinuous where:
- Rational functions have division by zero
- Piecewise functions change formulas
- Absolute value creates a corner (still continuous, but not differentiable!)
Why Continuity Matters
Continuous functions have nice properties:
- Intermediate Value Theorem: If continuous on [a, b], it takes every value between f(a) and f(b)
- Can find limits by direct substitution
- Behave predictably - no surprises!
Visual Test
On a graph, a function is continuous at a point if:
- No gap (both sides connect)
- No hole (no open circle)
- No jump (no sudden leap)
- The point is actually on the curve
📚 Practice Problems
1Problem 1medium
❓ Question:
Determine if is continuous at x = 3.
💡 Show Solution
Check 1: Does f(3) exist?
This is undefined! ✗
Conclusion: f is NOT continuous at x = 3 because f(3) doesn't exist.
The function has a hole at x = 3.
Note: Even though we can find the limit:
The limit exists (equals 6), but f(3) doesn't exist, so the function fails condition 1 of continuity.
2Problem 2hard
❓ Question:
For what value of k is continuous at x = 2?
💡 Show Solution
For f to be continuous at x = 2, we need:
Find f(2): Since , we use the second piece:
Find the limit from the left:
Find the limit from the right:
For continuity, these must all be equal:
Answer:
With this value, f(2) = 4 and both one-sided limits equal 4, making the function continuous!
3Problem 3easy
❓ Question:
Is f(x) = x² - 3x + 5 continuous at x = 2?
💡 Show Solution
Step 1: Check the three conditions for continuity at x = a:
- f(a) is defined
- lim(x→a) f(x) exists
- lim(x→a) f(x) = f(a)
Step 2: Check if f(2) is defined: f(2) = (2)² - 3(2) + 5 = 4 - 6 + 5 = 3 ✓ f(2) is defined
Step 3: Find lim(x→2) f(x): Since this is a polynomial, use direct substitution lim(x→2) (x² - 3x + 5) = 3 ✓ Limit exists
Step 4: Compare: lim(x→2) f(x) = 3 = f(2) ✓ All three conditions satisfied
Answer: Yes, f(x) is continuous at x = 2
4Problem 4medium
❓ Question:
Determine if f(x) = { x + 1, if x < 3; 5, if x = 3; 2x - 1, if x > 3 } is continuous at x = 3.
💡 Show Solution
Step 1: Check if f(3) is defined: f(3) = 5 ✓
Step 2: Find lim(x→3⁻) f(x): For x < 3, use f(x) = x + 1 lim(x→3⁻) (x + 1) = 3 + 1 = 4
Step 3: Find lim(x→3⁺) f(x): For x > 3, use f(x) = 2x - 1 lim(x→3⁺) (2x - 1) = 2(3) - 1 = 5
Step 4: Check if two-sided limit exists: lim(x→3⁻) f(x) = 4 lim(x→3⁺) f(x) = 5 Since 4 ≠ 5, lim(x→3) f(x) does not exist ✗
Step 5: Conclusion: Since the limit doesn't exist, f is not continuous at x = 3
Answer: No, f(x) is not continuous at x = 3
5Problem 5hard
❓ Question:
Find the value of k that makes f(x) = { (x² - 4)/(x - 2), if x ≠ 2; k, if x = 2 } continuous at x = 2.
💡 Show Solution
Step 1: For continuity at x = 2, need: lim(x→2) f(x) = f(2) = k
Step 2: Find lim(x→2) (x² - 4)/(x - 2): Direct substitution gives 0/0 (indeterminate)
Step 3: Factor numerator: x² - 4 = (x - 2)(x + 2)
Step 4: Simplify: (x² - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2) = x + 2 (for x ≠ 2)
Step 5: Evaluate limit: lim(x→2) (x + 2) = 2 + 2 = 4
Step 6: Set equal to f(2): k = 4
Step 7: Verify: With k = 4, lim(x→2) f(x) = 4 = f(2) ✓
Answer: k = 4
📋 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: What is Continuity?
Avoid these 4 frequent errors
🌍 Real-World Applications: What is Continuity?
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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