OK, in this problem, we're going to prove that L is not the limit. So we're going to use our definition F of X the absolute value minus the L less than the epsilon for X -, X not less than the delta. And what's going to really happen if we're trying to prove it's not the limit, we're going to actually show that the absolute value of F of X -, l has got to be greater than some epsilon or equal to. So the first thing we're going to do is we're going to realize we have two different pieces here. We're going to have a piece that says X - 7 greater than or equal to 1/2 when X is less than 4. And we're also going to have the absolute value of X + 3 - 7 greater than or equal to 1/2 for X greater than four. That's just the F of X -, l greater than or equal to epsilon. If we can show that those are true, it can't be greater than or equal to epsilon and less than epsilon at the same time. So the X -, 7 absolute value of X - 7 greater than or equal to 1/2 goes right there. Now it tells us that we want to solve this inequality. So I'm going to have X - 7 greater than or equal to 1/2, or the opposite of X - 7 greater than or equal to 1/2. That comes about by the official definition of absolute value. The absolute value of X says it's plain old X if X is greater than or equal to 0. But it's the opposite of X if X is less than 0. So when we solve this, now we know that this is true only for X values less than 4, so only on X less than four. That's important. So when we solve this, we get X greater than or equal to 7 1/2 or X - 7 if we divide it three by a negative less than or equal to negative 1/2, so X less than or equal to 6 1/2. So if we thought about looking at this on a number line, we'd have X less than or equal to 6 1/2 and X greater than or equal to 7 1/2. But it's only true for X less than four. So when we look at this, we have to think about it's only true. The original function was only true if X was less than four. So when we're filling in the second part, we need it to be X less than four because that's where they're both true when we're solving it. So the inequality is true for X less than 4. For this next piece, we're going to do the same thing. We're going to have, I'm going to simplify it up. So X -, 4 greater than or equal to 1/2, opposite of X -, 4 greater than or equal to 1/2. And in this case, it's for X greater than 4. So we have X greater than or equal to four and a half, and we have X -, 4 less than or equal to negative 1/2. So X less than or equal to 3 1/2. If we think about this one on a number line, we have X greater than 4 1/2 greater than or equal to. We have X less than or equal to 3 1/2. Now this one has to be true on the interval of X greater than four. So if I put a greater than 4 here, where are those both true? Those are both true from 4 1/2 on, so X has to be greater than nine halves, which is 4 1/2. The last part of this problem talks about the inequality is true for X greater than or equal to 9 halves. That's what we just showed. So for the function and epsilon being half, do there exist values of X for all delta greater than 0 such that F of X - 7 is greater than or equal to epsilon? Well, we found values in both of those instances in both of those cases, and for zero less than the absolute value X -, 4 less than delta. So the answer has to be yes, because we found X values that were true in both of those cases.