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AP Calculus BC 1.10 Exploring Types of 3 Discontinuities Study Notes

AP Calculus BC 1.10 Exploring Types of 3 Discontinuities Study Notes - New Syllabus

AP Calculus BC 1.10 Exploring Types of 3 Discontinuities Study Notes- New syllabus

AP Calculus BC 1.10 Exploring Types of 3 Discontinuities Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Exploring Types of Discontinuities

Key Concepts: 

  • Exploring Types of Discontinuities

AP Calculus BC-Concise Summary Notes- All Topics

Exploring Types of Discontinuities

Exploring Types of Discontinuities

 A function is continuous at \( x = a \) if all three conditions hold: 1. \( f(a) \) is defined. 2. \( \lim_{x \to a} f(x) \) exists. 3. \( \lim_{x \to a} f(x) = f(a) \). If any of these fail, the function is discontinuous at $x = a$.

Main Types of Discontinuities:  

1. Removable Discontinuity

  • Occurs when \( \lim_{x \to a} f(x) \) exists, but \( f(a) \) is either:
    • Not defined, or
    • Defined but not equal to the limit.
  • On the graph: A “hole” at \( x = a \).
  • Reason: Often caused by a common factor canceling out in rational functions.

Example: \( f(x) = \dfrac{x^2 – 1}{x – 1}, \; x \neq 1 \) Here: \( x^2 – 1 = (x – 1)(x + 1), \text{ so } f(x) = x + 1 \text{ for } x \neq 1. \) Thus: \( \lim_{x \to 1} f(x) = 2, \text{ but } f(1) \text{ undefined.} \)

2. Jump Discontinuity

  • Occurs when both one-sided limits exist but are not equal: \( \lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x). \)
  • On the graph: A jump (a gap between two points).
  • Reason: Typically appears in piecewise functions.

Example: \( f(x) = \begin{cases} 2x + 1, & x < 0 \\ x – 3, & x \ge 0 \end{cases} \) Here: \( \lim_{x \to 0^-} f(x) = 1, \quad \lim_{x \to 0^+} f(x) = -3. \) Different values → jump discontinuity at \( x = 0 \).

3. Infinite Discontinuity

  • Occurs when function values become unbounded near \( x = a \): \( \lim_{x \to a^-} f(x) = \pm \infty \text{ or } \lim_{x \to a^+} f(x) = \pm \infty. \)
  • On the graph: A vertical asymptote.
  • Reason: Usually caused by a zero in the denominator of a rational function.

Example: \( f(x) = \dfrac{1}{x – 2}. \) As \( x \to 2 \), \( f(x) \to \pm \infty \text{ (vertical asymptote at } x = 2\text{).} \)

Example: Classify the discontinuity of:

\( f(x) = \begin{cases} \dfrac{x^2 – 4}{x – 2}, & x \neq 2 \\ 7, & x = 2 \end{cases} \)

▶️ Answer/Explanation

Step 1: Simplify: \( \dfrac{x^2 – 4}{x – 2} = \dfrac{(x – 2)(x + 2)}{x – 2} = x + 2. \)

Step 2: Limit as \( x \to 2 \): \( \lim_{x \to 2} f(x) = 4. \)

Step 3: \( f(2) = 7 \). So limit exists but differs from \( f(2) \) → Removable Discontinuity at \( x = 2 \).

Example :

Identify and classify the discontinuity of the function:

\( f(x) = \begin{cases} 2x + 1, & x < 0 \\ x – 3, & x \ge 0 \end{cases} \)

▶️ Answer/Explanation

Step 1: Compute left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x + 1) = 1. \)

Step 2: Compute right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x – 3) = -3. \)

Since \( 1 \neq -3 \), \( \lim_{x \to 0} f(x) \text{ does not exist.} \)

Jump Discontinuity at \( x = 0 \).

Example: The graph of the function \( f(x) \) is shown. Which of the following statements is true about \( f \)?

  1. \( f \) is undefined at \( x = 1 \).
  2. \( f \) is defined but not continuous at \( x = 2 \).
  3. \( f \) is defined and continuous at \( x = 3 \).

(A) Only I    (B) Only II    (C) I and II    (D) I and III    (E) None of the statements are true.

▶️ Answer/Explanation

Step 1: At \( x = 1 \), the graph shows an open circle at \( y = 2 \) and a filled dot at \( y = 3 \). This means \( f(1) = 3 \), so the function is defined at \( x = 1 \).
 Statement I is FALSE.

Step 2: At \( x = 2 \), there is a filled circle and the left and right limits match, so the function is continuous at \( x = 2 \).
Statement II is FALSE.

Step 3: At \( x = 3 \), the graph has a vertical asymptote, so \( f(3) \) does not exist.
 Statement III is FALSE.

Correct Answer: (E) None of the statements are true.

Example :

Identify and classify the discontinuity of the function:

\( f(x) = \dfrac{1}{x – 2} \)

▶️ Answer/Explanation

Step 1: Observe the denominator: When \( x = 2 \), denominator becomes zero → potential asymptote.

Step 2: Compute one-sided limits: \( \lim_{x \to 2^-} f(x) = -\infty, \quad \lim_{x \to 2^+} f(x) = +\infty. \)

The function becomes unbounded near \( x = 2 \). Infinite Discontinuity at \( x = 2 \).

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