AP Calculus AB· difficulty 2/5
From the graph, exists because the left and right limits agree (open circle), even though (filled circle) is different. The function is
Discontinuous at — infinite
Discontinuous at — removable
Discontinuous at — jump
Continuous at
Explanation
The limit exists but does not equal . Redefining to the common value of the limits would make continuous — that's a <strong>removable</strong> discontinuity.
Want 10 more like this — adaptive to your weak spots?
The graph shows f with \lim{x\to 2^-}f(x) = 3, \lim{x\to 2^+}f(x) = 5, and f(2) = 5. The discontinuity at x = 2 is…
chevron_rightThe 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…
chevron_rightFrom the graph, the open circle at (1, 3) shows that \displaystyle\lim{x\to 1} f(x) =…
chevron_right