# Why don't these limits for the piecewise function and graph make sense?

I've followed my notes and examples in my text book as well as online. My reasoning for the answers I came up with seem to make sense by looking at the graph, but I must be doing something wrong.

I've convinced myself several times and have tried DNE,0,1, and 2 for each one.. I cannot figure it out.

It's a fairly basic problem.. so I took screenshots of the question with my answers and the graph (via wolfram).

Question: Graph: • Is it expecting all three results to be correct before it accepts it? I would recommend figuring out what the correct values are for all three, and then try entering them. – Amzoti Sep 13 '13 at 4:39
• What amzoti (and zodiac below) are getting at is that your answer to #2 is wrong... What is the "right-hand limit" of your function? – colormegone Sep 13 '13 at 4:43
• So would it be 0,2,2? – Cozen Sep 13 '13 at 4:47
• Your first two answers are now correct. But if the "left-hand" and "right-hand" limits at a point do not agree, then the "two-sided" limit does not exist. The two-sided limit is what is being asked about in #3. – colormegone Sep 13 '13 at 7:42

$\lim_{x \to -2^+} f(x)$ means the limit of $f(x)$ as $x$ approaches $-2$ from the right, for example, $-1.9, -1.99, -1.999 \ldots$ and should be $2$.

By the way, your answer as it stands is inconsistent; if both the left and right limits exist and are equal, then the limit exists by definition.

Edit: You seem confused about the two-sided limit. The limit

$$\lim_{x \to a} f(x)$$

exists only when both the below limits exist and are equal to each other

$$\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x)$$

• I just tried 2 again and it says it's not correct – Cozen Sep 13 '13 at 4:30
• Did you change your answer for the third question as well? Remember that the answer to (3) depends on the answer to (1) and (2). Also remember that the limit in (3) is not the same as $f(2)$ – zodiac Sep 13 '13 at 4:38