Example exercise using limit flow chart

In a previous video, we introduced this flowchart that helps us think about what strategies to use when trying to determine a limit of a function as it approaches a point. What we're going to do in this video is now try to apply that in several example exercises.

So, the first question they say, based on the process in the flowchart, let's try to find the limit as x approaches 3 of f of x for f of x is equal to this business. What are the steps that we should go through? They have these various steps. You might say, what is a, b, c, d, and a? Well, if you go back to the original flow chart, you see this is a, this is b, this is c, this is d, etcetera, etcetera.

So, let's just first try to substitute, see if we can get a value for f of 3. If f of 3 works, then we are essentially done because this is not some strange exotic function; it seems like it's continuous if function around x equals 3, if f 3 exists. So, let's try it out.

So, we're going to have f of 3 is equal to the square root of, let's see, 6 minus 5, 2 times 3 minus 5 minus 1 over 3 minus 3. In the numerator here, we get 0 because this is 1 minus 1. So this is going to be equal to 0 over 0. The first thing that happens is we fall into indeterminate form.

So, the first step is going to be c. Now, the next thing that I would do, because we have a radical expression here, is to see if I can take a conjugate. So, it's going to likely be we're definitely going to have c first, then we're going to try e, and frankly, that's the only choice that has that. But let's actually see if that actually works: c, e, b.

So, let me multiply this expression times a conjugate, so square root of 2x minus 5 minus 1 over x minus 3 times square root of 2x minus 5 plus 1. That's the conjugate here over square root of 2x minus 5 plus 1.

This is going to be equal to see this numerator. Square root of 2x minus 5 squared is going to be 2x minus 5, and then you're going to have my negative; you have a minus 1 squared, which is going to be minus 1. All of that over x minus 3 times square root of 2x minus 5 plus 1.

Let's see this stuff up here. You can rewrite as 2x minus 6, which is the same thing as 2 times x minus 3, and so those cancel out nicely. So, let's see, this is going to be equal to 2 over square root of 2x minus 5 plus 1.

Now, let's try to evaluate this when x equals 3. Well, if we do that, we're going to get 2 over, so this is going to be 6 minus 5, the square root, which is 1 plus 1. So it's 2 over 2, which is equal to 1.

So then, we were able to evaluate it, and we actually got a value. So, we feel pretty good that we're done. We went from finding indeterminate form, taking the conjugate, and then being able to evaluate it.

Let me just show that right over here. When we immediately tried to evaluate it, we got indeterminate form. We recognized that it was a radical, that it was a rational expression with a radical on top. So, hey, let's multiply by the conjugate. We did that, and then after we were able to simplify it with the conjugate, we tried to evaluate it again, and we were able to get a real number.

So, we feel pretty good that that is going to be the limit.