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Answer:
a. f(0) = 1
b. DNE (does not exist)
c. DNE
d. lim = 3
Step-by-step explanation:
The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.
The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.
This discussion tells you ...
f(0) = 1
f(2) does not exist. There is a "hole" in the function definition there.
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The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.
Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.
Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.
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Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."
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In summary, ...
a) f(0) = 1
b) lim x → 0 does not exist
c) f(2) does not exist
d) lim x → 2 = 3
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<em>Additional comment</em>
The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").
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