 Calculus is based on the idea of a limit, so here's a quick introduction to what limits are. So, we might begin by considering the limit of a polynomial function. For example, here we have the limit as x approaches 2 of x squared minus 4x plus 5. Now, there's actually two parts to this problem. One is to actually find this limit, but the other thing is we want to defend the answer numerically. Come up with some evidence that our claim is actually correct. So what's the idea behind a limit? In this particular case, what we're interested in finding out is what happens as x gets close to 2. If you want to read this expression, it's really asking us what happens to this thing, x squared minus 4x plus 5, as x gets close to 2. So, one way we might proceed with this is that as x gets close to 2, x squared gets close to 2, squared 4x gets close to 4 times 2, and 5 gets close to... well, it actually gets close to 5, because it's always 5. So as x gets close to 2, this polynomial expression gets close to 2 squared minus 4 times 2 plus 5, and that simplifies down to 1. And so we can say that our limit as x approaches 2 of 4x squared minus 4x plus 5 is equal to 1. As x gets close to 2, it looks like this thing gets close to 1. As an answer to the first part of the question, we don't really have to do anything else. However, we also want to defend our answer numerically. So let's see what actually happens as x gets close to 2. And so we might do that by constructing a table of values for our values of x close to 2, and our expression x squared minus 4x plus 5. So we'll pick some x values close to 2. For example, if x is 1.9, then I'll substitute that into the expression and get 1.01 as the result. Well, let's get even closer to 2. I might take a look at x equals 1.99, and again, if I substitute 1.99 into the expression, I'm going to end up with something that looks like this, 1.001. And close to 2 could be above it, so maybe I'll take 2.1 and find out what the expression is equal to, or 2.01, and find out what the expression is equal to. And looking at these values here, as x gets close to 2, it does look like the values of this expression are getting close to 1. So I might say that the values in the table support the claim that as x gets close to 2, this expression gets close to 1. And if I wanted to write a complete answer to this entire question, and find the limit and defend the answer numerically, I need to include everything that's in green. Well, I can also take a look at a rational function, find the limit as x approaches 3, 3x plus 5 over x squared plus 2x minus 2. In other words, what happens to this expression as x gets close to 3, and as before, we'll also take a look at defending our answer numerically. So what happens? Well, as x gets close to 3, we might analyze the problem as follows. This numerator expression, 3x plus 5, looks like it's going to get close to 3 times 3 plus 5. That's 9 plus 5, that's 14. Our denominator, x squared plus 2x minus 2, looks like it's going to get close to 3 squared plus 2 times 3 minus 2. And I'll go ahead and simplify that. That works out to be 13. And if I want to consider then the quotient 3x plus 5 over x squared plus 2x minus 2, then it seems reasonable that that quotient is going to get close to the value 14 over 13. And so I may make the claim that by limit as x gets close to 3 of this thing, I'm guessing it's going to be about 14 thirds. Well, let's see if we can find that answer numerically. So we'll check out the values of this expression near x equal to 3. So again, we'll construct a table. And so let's pick an x value close to 3. How about 2.9? I'll substitute that into my expression. And after all the dust clears, I get a numerical value 1.1220 approximately. And I'll take a closer value 2.99. I'm close to 3. Substitute that in. And all the dust clears. I end up with 1.0813 about. And again, I don't have to only consider values less than 3. I should consider values greater than 3. So how about 3.1? And substitute that in. Let all the dust clear. And something even closer to 3. And I get something like that. And so the question is, do the table values here support the claim that my limit is going to be 14 over 13? Well, it helps if I knew what 14 over 13 was as a decimal. So I'll go ahead and compute that. That's about 1.0769. And it does seem that as I get close to 3, if I'm close to 3, the values here are pretty close to this 1.0769. So it seems reasonable to conclude that this work here is correct. And so my values support the claim that the limit is 14 over 13. And again, as a complete answer to the question, find this limit and defend the answer numerically, I'm going to include everything that's in green. And for a radical function, I can do essentially the same sort of problem. So here, what happens to square root x plus 5 as x gets close to negative 1? And the same analysis goes through as x gets close to negative 1. This square root x plus 5 gets close to negative 1 plus 5. And after all the dust settles, looks like that's going to get close to 2. And my guess is that my limit is going to be 2. But again, let's take a look at our answer numerically. We'll try and defend this claim by computing values. So we'll take a look at square root of x plus 5. That's for x values that are close to negative 1. And again, I'll set up a table. Here's something close to negative 1. Works out to be square root 3.9, 1.9748. And take something closer. And after all the dust settles, 1.9975. And again, I don't have to be on one side. I can be on the other side. And again, I'll do the computation. And again, take something even closer. And let's take a look at it. If I take a look at x values that are close to negative 1, it does look like square root of x plus 5 is close to 2. And so again, as a complete answer to the question, find the limit and defend the answer numerically, you do want to include everything that's written in green.