>> Here is another one. This is interesting because we're not finding the limit of a function so much as we're finding the limit of an expression that is written in terms of a function. And the function is this little simple function. And the expression turns out to be the difference quotient. We studied difference quotient in an earlier chapter when we were first talking about functions generally and function notation. And here, we're reemphasizing the idea of difference quotient. Difference quotient comes up quite often in calculus, in the early studies of calculus. And we may have a chance to see it a little bit more in this course as well. But here is how we take care of the situation. We recognize this. Upon replacing H with 0, we recognize this as indeterminate form. Because after all, if H is 0, we would have F of X here, then minus F of X. That's 0 in the numerator. And then when H is 0 here, 0. So 0 over 0, indeterminate form. That just tells us that we need to go through some kind of manipulation. The manipulation here is just the simplification of this expression using the information about the function itself. Now, all we're doing here is using what we know about function notation and what it means. So I have written the function up here. I just repeated it just so that we can look back at it a couple of times. Here is our difference quotient expression. And now, what does it mean? Well, let's see, F of X plus H means in the F function, replace X with X plus H. So here is the F function, and here is the expression for it. And we're replacing the X with X plus H. So we have 3 times X plus H minus 1 for this. Now, for emphasis, I want to separate the idea of this part of the numerator and this part of the numerator. And I'm doing that with symbols of inclusion somehow. So in the brackets here, F of X plus H is this, all right? All it is is this expression with X replaced with X plus H. That's all there is to it. Okay, then minus F of X means minus this. So it's minus, here is F of X, you see? And now we go through and simplify, we have 3 times X, 3X. 3 times H, 3H. Bring down the minus 1, this is minus 3X plus 1. And now, a lot of things go out very quickly here. 3X minus 3X go out. We have minus 1 plus 1. They go out. So we have 3H over H for our fraction. The Hs now go out, and we have a constant 3. So the situation becomes relatively trivial. Now we find the limit as H approaches 0 of the constant 3, and we know that that is simply the constant.
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