The function f(x) = \dfrac{x^2 - 9}{x - 3} has what kind of discontinuity at x = 3?

Answer: Removable (hole) Factor: f(x) = (x-3)(x+3)/(x-3) = x+3 for x \ne 3. The function has a removable hole at x = 3, where it can be redefined as f(3) = 6.

Which of the following is NOT continuous at x = 0?

Answer: h(x) = 1/x 1/x is undefined at x = 0 (vertical asymptote). The others are continuous everywhere.

The graph has a hole at the peak. The limit at that x-value

Answer: Is the open-circle value (top of curve) Hole → function undefined there, but the limit equals the value the curve approaches.

AP Calculus AB · Topic 1.11

Removing Discontinuities Practice

Part of Limits and Continuity.(LIM-2.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$f(x) = \dfrac{x^2 - 9}{x - 3}$
- A
  $Removable (hole)$
  check_circle
- B
  $Jump$
- C
  $None — it is continuous$
- D
  $Infinite (asymptote)$
Why
$f(x) = (x-3)(x+3)/(x-3) = x+3$
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Sample 2difficulty 2/5
$x = 0$
- A
  $f(x) = x^2 + 1$
- B
  $j(x) = e^x$
- C
  $h(x) = 1/x$
  check_circle
- D
  $g(x) = \sin x$
Why
$1/x$
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Sample 3difficulty 2/5
$x$
- A
  $Is the open-circle value (top of curve)$
  check_circle
- B
  $Equals zero$
- C
  $Doesn't exist$
- D
  $Is undefined$
Why
$Hole \to function undefined there, but the limit equals the value the curve approaches.$
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