AP Calculus BC 1.11 Defining Continuity at a Point Study Notes - New Syllabus
AP Calculus BC 1.11 Defining Continuity at a Point Study Notes- New syllabus
AP Calculus BC 1.11 Defining Continuity at a Point Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
Defining continuity at a point.
Key Concepts:
- A function \( f(x) \) is continuous at \( x = a \) if:
- \( f(a) \) is defined
- \( \displaystyle \lim_{x \to a} f(x) \) exists
- \( \displaystyle \lim_{x \to a} f(x) = f(a) \)
Defining Continuity at a Point
Defining Continuity at a Point
A function \( f(x) \) is said to be continuous at a point \( x = a \) if the following three conditions hold:
- \( f(a) \) exists (the function is defined at \( x = a \)).
- \( \lim_{x \to a} f(x) \) exists (the limit exists from both sides).
- \( \lim_{x \to a} f(x) = f(a) \) (the limit equals the function value).
Mathematical Representation:
\(\text{f is continuous at } x = a \iff \lim_{x \to a} f(x) = f(a)\)
Key Idea: If any one of these conditions fails, the function is discontinuous at that point.
Types of Discontinuity:
- Removable Discontinuity: Limit exists but \( f(a) \) is undefined or different.
- Jump Discontinuity: Left-hand and right-hand limits exist but are not equal.
- Infinite Discontinuity: Function approaches \( \pm \infty \) near the point.
Example:
Check continuity of \( f(x) = x^2 \) at \( x = 2 \).
▶️ Answer/Explanation
Step 1: Compute \( f(2) \):
\( f(2) = (2)^2 = 4 \)
Step 2: Compute \( \lim_{x \to 2} f(x) \):
\( \lim_{x \to 2} x^2 = 4 \)
Step 3: Compare:
\( \lim_{x \to 2} f(x) = 4 \) and \( f(2) = 4 \)
All conditions satisfied, so \( f(x) \) is continuous at \( x = 2 \).
Example:
Check continuity of \( f(x) = \dfrac{x^2 – 4}{x – 2} \) at \( x = 2 \).
▶️ Answer/Explanation
Step 1: Check \( f(2) \):
\( f(2) = \dfrac{4 – 4}{0} \) → Undefined.
Step 2: Simplify and compute limit:
\( f(x) = \dfrac{(x – 2)(x + 2)}{x – 2} = x + 2 \) for \( x \neq 2 \)
\( \lim_{x \to 2} f(x) = 2 + 2 = 4 \)
Step 3: Compare:
Limit exists (4), but \( f(2) \) is undefined.
Function is not continuous at \( x = 2 \) (Removable discontinuity).
Example:
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} kx^2 + 1, & x < 2 \\ 3x – k, & x \geq 2 \end{cases} \]
Find the value of \( k \) so that \( f(x) \) is continuous at \( x = 2 \).
▶️ Answer/Explanation
For continuity at \( x = 2 \), we require:
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \).
Compute Left-hand limit (LHL):
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2 + 1) = k(2)^2 + 1 = 4k + 1 \).
: Compute Right-hand limit (RHL):
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x – k) = 3(2) – k = 6 – k \).
For continuity, LHL = RHL = f(2):
\( 4k + 1 = 6 – k \).
Step 5: Solve for \( k \):
\( 4k + k = 6 – 1 \)
\( 5k = 5 \)
\( k = 1 \).
Final Answer: \( k = 1 \). The function is continuous at \( x = 2 \) when \( k = 1 \).
Example:
The function \( f \) has the properties indicated in the table below. Which of the following must be true?
| Property | b = 1 | b = 2 | b = 3 |
|---|---|---|---|
| \( \lim_{x \to b^-} f(x) \) | -1 | 5 | 1 |
| \( \lim_{x \to b^+} f(x) \) | 3 | 5 | 1 |
| \( f(b) \) | 3 | 8 | 1 |
(A) \( f \) is continuous at \( x = 1 \).
(B) \( f \) is continuous at \( x = 2 \).
(C) \( f \) is continuous at \( x = 3 \).
(D) None of the above.
▶️ Answer/Explanation
Step 1: A function is continuous at \( x = b \) if:
\( \lim_{x \to b^-} f(x) = \lim_{x \to b^+} f(x) = f(b) \).
For \( x = 1 \): LHL = -1, RHL = 3, f(1) = 3 → LHL ≠ RHL → Not continuous.
For \( x = 2 \): LHL = 5, RHL = 5, so limit = 5. But f(2) = 8 → Not continuous.
For \( x = 3 \): LHL = 1, RHL = 1, so limit = 1, and f(3) = 1 → Continuous at \( x = 3 \).
Answer: (C) \( f \) is continuous at \( x = 3 \).
Example:
Check continuity of \( f(x) = \begin{cases} x^2, & x < 1 \\ 3, & x = 1 \\ x + 1, & x > 1 \end{cases} \) at \( x = 1 \).
▶️ Answer/Explanation
Step 1: \( f(1) = 3 \)
Step 2: Compute LHL and RHL:
\( \lim_{x \to 1^-} f(x) = 1^2 = 1 \)
\( \lim_{x \to 1^+} f(x) = (1) + 1 = 2 \)
Since LHL ≠ RHL, the two-sided limit does not exist.
Function is not continuous at \( x = 1 \) (Jump discontinuity).