The graph has a vertical asymptote at x = 1 with the function going to +\infty on the right and -\infty... wait, see the graph: left side goes to -\infty, right side from +\infty. What is \lim{x\to 1}

Answer: Does not exist Left and right one-sided limits diverge to opposite infinities, so the two-sided limit does not exist (and is not \pm\infty).

From the graph, the open circle at (1, 3) shows that \displaystyle\lim{x\to 1} f(x) =

Answer: 3 Both one-sided limits approach 3 even though f(1) is not defined there.

From the graph, \displaystyle\lim{x\to 2} f(x) exists because the left and right limits agree (open circle), even though f(2) (filled circle) is different. The function is

Answer: Discontinuous at x = 2 — removable The limit exists but does not equal f(2). Redefining f(2) to the common value of the limits would make f continuous — that's a <strongremovable</strong discontinuity.

AP Calculus AB · Topic 1.3

Estimating Limit Values from Graphs Practice

Part of Limits and Continuity.(LIM-1.C)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$x = 1$
- A
  $+\infty$
- B
  $Does not exist$
  check_circle
- C
  $-\infty$
- D
  $0$
Why
$\pm\infty$
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Sample 2difficulty 2/5
$(1, 3)$
- A
  $1$
- B
  $0$
- C
  $3$
  check_circle
- D
  $Does not exist$
Why
$3$
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Sample 3difficulty 2/5
$\displaystyle\lim_{x\to 2} f(x)$
- A
  $x = 2$
- B
  $x = 2$
  check_circle
- C
  $x = 2$
- D
  $x = 2$
Why
$f(2)$
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Sample 4difficulty 2/5
$f$
- A
  $Removable$
- B
  $None$
- C
  $Jump$
  check_circle
- D
  $Infinite$
Why
$Left and right limits are different finite values \to jump discontinuity.$
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