Question

Review the graph of function f(x). On a coordinate plane, a graph has maximum point at open circle (0, 4), and curves down and then repeats with smaller curves at intervals of 2 pi in both the positive and negative x -axis. A point is at (0, negative 2). Which statement identifies and explains Limit of f (x) as x approaches 0? The limit Limit of f (x) as x approaches 0 = –2 because the value of the function at x = 0 is –2. The limit Limit of f (x) as x approaches 0 does not exist because there is an open circle at (0, 4). The limit Limit of f (x) as x approaches 0 = 4 because both the left-hand and right-hand limits equal 4. The limit Limit of f (x) as x approaches 0 does not exist because there is oscillating behavior around x = 0.

Answer 1

The true statement is that (b) The limit $\lim_{x \to 0} f(x)$ does not exist because there is an open circle at (0, 4).

How to interpret the limit of f(x)?

From the graph of the function (see attachment), we have the following highlight:

There is an open circle at point (0,4), the maximum point of the function f(x) and when x = 0

When there is an open circle in the graph at x = 0, then the function has no limit at this point

This is so because the value is exclusive of the function values

Hence, the limit $\lim_{x \to 0} f(x)$ does not exist because there is an open circle at (0, 4).

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