All right. Hi everyone. Let's do a couple of more examples using the limit definition of the derivative. Whenever we do one of these problems, let's just write the formula down, tells the teacher you know the formula, catch some points in case everything else goes horribly wrong. So remember the set-up here. I'll give you some function F of X. I pick a point A, and I am interested in finding the slope of the tangent line at A. What is this thing? This slope over here will be our derivative at A and our variable is going to be H. We're going to think of H as the distance away from A, and we're going to let H go to zero in the limit. So here's our limit definition of the derivative limit as H goes to zero of F of a plus h minus f of a, all over h. There is a another way. So I'll say version number two, there's another way to think of this, and it comes up once in a while. If you make the substitution x equals a plus h and then solve for H to get x minus a. You can also see it as follows. So f prime of a is now the limit; as X approaches A of f of x minus f of a, all over x minus a. These two are equivalent, they're equivalent definitions. So you could see either one. For the most part, we're going to use the one at the top, but I want you to know both. We'll do some examples where this form is actually a little more handy. So have these two down, have these two ready to go and we'll do an example to just get more practice. So here we go. I think I left it hanging on the last one, so let's do that one. F of X equals one over x squared at x equals one. So one over x squared, that's our volcano function, looks something like this. We have a point at one. So one over one squared is just one. There is a very nice tangent line at this point and it has a slope of course. So find the slope, well, this would be the derivative plugged in at one. How do we find this line? Well, let's use the limit definition of the derivative. So here we go. So what are we after it? Let's write the notation F prime of one is the limit as H goes to zero. Let's use the first definition of F of one plus h minus f of one, all over h. So I write the limit definition down. Now by the way, make sure you have limits here; once in a while I see students leave this off and it breaks my heart because it's the limit definition of the derivative. So the word limit is built in and if you leave it off it again, you're missing the point here of this entire chapter. So f of one plus h. So let's fill it in with our particular function, with our particular value and we can work this out. So the function is, take the reciprocal and square this thing. So we have one over one plus h squared minus, we said f of one is one, so just minus one, all of that over h. I have a nice big growth function. But hopefully we know how to handle this. This thing screams to clean me up, simplify that numerator. So if you one is like one over one. When you do this and put overcomes denominator, you get one minus one over one plus h squared over one plus h whole thing squared. There's an h on the bottom which stays on the bottom so I'll just throw it over here. I'm skipping some algebra, so if you need to pause the video and do some algebra, please do so to work that out. But this complex fraction I should say, cleans up to be this thing. So now what should I do? Well, there's, let's see, what should we do? The numerator is screaming to foil this thing. So you have one minus use parentheses, one plus 2h plus h squared over h one plus h squared. Now notice I had to like keep going. Say why can't I plug in? Well, there's still this h on the bottom, on the denominator. I can't plug in zero downstairs. It hasn't canceled with anything yet. So let's keep going and hopefully something will cancel to the point where I can plug in. We distribute the minus sign. You get one minus one. So 1 minus 1 minus 2h minus h squared, over h1 plus h squared. This is promising. I think I see something that's going to cancel. If you notice the ones go away immediately. One minus one is just zero. So they go away, bye-bye. Then there's an h in each term that's left. So you can imagine factoring that out to get minus two minus h over the h one plus h, quantity squared. I'll go back to my other color. Well that's nice. There go the H's. Alright so now that problem in the denominator has just resolved itself. So the limit as H goes to zero, what am I left with? Minus 2 minus h, over 1 plus h squared. This H downstairs is not a problem because when I plug in, which I'm allowed to do now, and again, notice the limit has actually gone away because I'm evaluating the limit. You get one plus zero. That is not division by zero, that's division by one squared. This is perfect, because I'm left with minus 2 minus 0, that's just minus 2 over 1, and that's just better known as minus 2. This slope of the line is minus two, I was expecting it to be negative, that's nice. So minus 2 is my final answer using limits. Let's do another example, just got to do more of these. Let's find the derivative at 3 when f of x is, we'll do this quadratic polynomial, x squared plus 5x plus 1, so we're going to use the limit definition of the derivative. Let's write it down, here we go. F prime of 3 is equal to the limit as h goes to 0 of f of 3 plus h, minus f of 3 all over h. Write the definition down, just keep doing it to let it sink in, help you memorize it as we go. Notice it got the limit, got equal signs, everything's very nice. Now I do the limit as h goes to 0, f of 3 plus h. Wherever I see an x, I plug in 3 plus h, I can do this. Use parentheses, 3 plus h squared plus 5 parentheses, 3 plus h plus 1. Everywhere I see an x, I plug in 3 plus h, great. Then I plug in the function evaluated at three, so let's just go off on the side here for a second. What is f of 3? This becomes 3 squared plus 15 plus one, it's 25, 9 plus one plus 15, 25, 10 plus 15, 25. So minus 25, that's the function evaluated at 3, all of this is over h. I can't plug in yet, that big [inaudible] is preventing me from doing that, so let's just try to clean up this numerator. What happens when I do that? Foil the top, you get 9 plus 6h plus h squared plus 15, plus 5h plus 1 minus 25, I guess I can just combine that. That would be what? Minus 24 all over h. If I'm going fast with the algebra and you need more time, that is perfectly fine. Just pause the video and catch up as you need. There's no wrong number of steps here, show the number of steps that you need. So what do I got? I have an h squared upstairs and then I have all my h terms, so I got that one. There's a 6h and a 5h, add those up, you get 11h. Then constants, we have 9 plus 15, that's 24, minus 24, that's 0. That's great, so they all cancel, 9, 15, and 24 go away and we'll divide by h. Similarly, seeing the pattern that's starting to emerge, there's an h in both terms. I'll factor out one h from each term in the numerator, allowing me to cancel with the h upstairs and I get the limit as h goes to 0 of h plus 11. How nice is that? Because now I plug in, so I don't need to write the limit here, because I'm about to take the limit, about to evaluate the limit, I get 0 plus 11. That's brilliant, carry the 12, divide by Pi, you get 11. The derivative, the slope of this line, this quadratic at three is positive 11. One more, and this will be, I'll put a star next to this one, this is el classico. This one is a tricky one because I'm going to give you a piecewise function. Of course, whenever you work with piecewise functions, it's a little more challenging to work with them just because you've got two pieces to treat versus one. Here's the function. It's a piecewise function, f of x equals x sine of 1 over x for x naught 0 and then add 0, let's just define it to be 0. Obviously, the top I can't plug in 0 because of that. This is some function, has a graph and the question is, what's going on with f prime of 0? What does this equal to? So let's use the limit definition to figure this out, ready? We have the limit as h goes to 0 of f of, and normally it's the number plus h. Our number here is 0 plus h. I'll just write it out, obviously I could have wrote h, minus f of 0 all over h. This is the generic formula, let's plug it in for our particular function. I'm going to write the limit every time. Do I have to write the limit every time? Yes, you have to write the limit every time. I'm going to evaluate this function now. This is a little more challenging, not hard, but just something to keep track of. Remember I have to use the piece where it's defined, f of 0 is equal to 0. So I want to just make sure that I'm evaluating my piecewise function correctly. This is minus 0, do I need to write that? Can you guys handle that? I'll write it, I feel bad. If h_1 is 0, that's good. Now here's a question for you. I want to evaluate this h. Now that we have to understand limits a little bit, remember h is approaching 0. Maybe it's like 1, one-half,one-third, one-fourth when it's getting there from both sides. The key thing about a limit remember as I approach a point, I'm not the point. I'm basically every number except 0. To evaluate f of h, I'm going to use the top branch of my piecewise function. This becomes the limit as h goes to 0. I'm going to plug in h, but up here to h sine of 1 over h. Where I see an x, I get that. Minus 0, that just goes away and then I'm over h. The derivative at 0 of this piecewise function is going to need to be whatever this limit turns out to be. The h is cancel always a good sign when things are canceling, we like that, and I'm left with this limit now. We've actually done this one before, but may not remember it. Sine of 1 over h. Now this is a tricky limit. What can I do with this one? Remember sine is a continuous function. By the theorem, it has no name. We can move the limit inside. Whenever you have a continuous function, move it inside. Limit of 1 over h. Now, limit of 1 over h, as h goes that seems like 1 over x. Remember that function drawn this picture 100 times. This is the graph of 1 over h. In this case, of course, 1 over x, same thing, dummy variable. What's the limit as I approach 0? As I approach from the right, I go to infinity, is approaching a left eye going to negative infinity. This limit does not exist. Sine of something that doesn't exist, of course this whole thing doesn't exist. Here is a graph of a function. Something is going on with this graph, we saw this one is a little compresses and oscillates like crazy. This does not have a well-defined derivative. This the first example that usually comes up. This is like classic of a function whose derivative does not exist at a point. Remember this can happen. Limits are part of the formula for derivatives and so not every limit exists and here's one that doesn't. Keep this in mind as we do it. Now, I'm going to show you one, let's make one adjustment. This one trade is done. I'm going to use a different color and make one adjustment. Maybe I'll do it in like crazy purple. What if I had started with x squared? I'm going to do it in more room just to show you what can also happen, how a slight change to a function can make a world of difference. What if I start with x squared? I'll ask you the same exact question, I promise you all. Everything is the same here. This step doesn't change. We plug in 0 plus h minus the function evaluated at 0 over h, that doesn't change, the step doesn't change. The only thing that changes here is now I have an h squared. There's an h given there, h squared. When I have an h squared, I still cancel one of the hs, but I'm left with an h back here. I'm left with an h in front of my sine of 1 over h. Now that changes things a lot because I can't quite bring in the function and do the rules on it because not defined. What changes now is this goes away. Let's work this out for the h squared, the x squared case. This is like example Number 2 with a slight variation. Again, this is why this is such a classic example because one little square can make or break this thing. Where did we leave off or trying to evaluate the limit as h goes to 0 of h of sine of 1 over h. I can't just plug in. That doesn't work anymore. We've also seen this one before too. They were setting up for this one. Do you remember how to do this one? If you think about it, I have something that's going to 0 and have a function that is bounded. A function going to 0, sum is bounded when you have 0, like going to 0 and the limit has a bounded function, you squeeze theorem. Sine is bounded above by 1, so I'm going to think of it as limit as h goes to 0 of h times 1, its upper limit of sine is just 1, and then bounded below by the limit as h goes to 0 of h times negative 1. I'm going to bound the sine function by its upper and lower limits. In that case, what happens, the limit as h goes to 0, let's just say that this thing goes to 0. This goes to negative 0, but it's of course is 0, the left side. I have some number that's between greater than 0 and less than or equal to 0, that goes to 0 must be 0 by the squeeze theorem. This is amazing. Because with just x by itself, the derivative doesn't exist. The limit doesn't exist. With an x squared, the derivative is 0. There's two examples here, and just by changing it to a square, using the squeeze theorem, I get the derivative to exist. But if I just keep it as x times sine of 1 over x, then this thing doesn't exist. That's amazing that this happens. Keep this one in mind. This is really two examples in one and make sure you go through both and understand both. This is a classic example of a derivative that doesn't exist for that point. Go over this and let me know if you have any questions.