There's a lot going on in the previous activity, so let's just focus on the word "[b]limit[/b]."[br][br]First of all, what is a [b]limit[/b] exactly?[br][br][b]Definition[/b]: A [b]limit[/b] is a [i]study[/i] of a non-terminating process or pattern that determines the most logical conclusion of that non-terminating process or pattern.[br][br]For instance, in the previous activities, we were [i]studying[/i] the slopes of the secant lines as [code]h[/code] approaches 0. This process is non-terminating because we can't set [code]h[/code] equal to 0 and see "the end", because doing so results in no line. So we need to study the [i]process[/i] when [code]h[/code] tends to 0, and determine what the outcome of this process is some other way. As we were able to see in the previous activity, the slope of the secant lines when [code]h[/code] approaches 0 displays an obvious pattern. Specifically, when [code]h[/code] trends to 0, the slope of [code]g[/code] trends to 4. [br][br]The mathematical way of saying this is "the limit of the slope of [code]g[/code] as [code]h[/code] tends to 0 is 4." The usual mathematical shorthand for this is:[br][br][math]\lim_{h\longrightarrow0}slope\left(g\right)=4[/math][br][br]This is more than enough for this book, but as a nod to a traditional calculus calculation, let's try to express [code]slope(g)[/code] algebraically, and see if we can get a deeper explanation of why this limit is equal to 4. If you are already convinced the limit is equal to 4 from the previous activity, this won't do much to improve your conviction. But if you feel like you need to see more of an explanation, that's what we'll do below. If you'd rather not, you can go ahead and skip the rest of this section, and you'll be just fine for the rest of the book.[br][br]Now, for those you who are still with me, let's work this algebra out step by step. Thinking back to the actives from earlier in this chapter, the line [code]g[/code] depends on two points, [code]A=(1,3)[/code], and a point close to [code]A[/code] that depended on the "nudging" variable [code]h[/code]. The way we made the other point and incorporated [code]h[/code] was by simply adding it to the x coordinate of [code]A[/code]. Indeed, this was the code for the second point:[br][br][code](x(A)+h,f(x(A)+h))[/code][br][br]The code [code]x(A)[/code] is a Geogebra tool that obtains the x coordinate of point [code]A[/code]. [br][br]We can use these two points to calculate the slope of [code]g[/code] directly. Specifically, the slope of [code]g[/code] is the difference in the y coordinates divided by the difference in the x coordinates ("rise over run"):[br][br][math]\frac{f\left(x\left(A\right)+h\right)-3}{x\left(A\right)+h-1}[/math][br][br]Because [code]x(A)[/code], the x coordinate of [code]A[/code], is 1, we can replace that as well. Notice this allows us to simplify the denominator (bottom of the fraction).[br][br][math]\frac{f\left(1+h\right)-3}{1+h-1}=\frac{f\left(1+h\right)-3}{h}[/math][br][br]This last expression is an algebraic expression for [code]slope(g)[/code] in the limit higher up on this page. So, we can swap [code]slope(g)[/code] out, and replace it with this expression:[br][br][math]\lim_{h\longrightarrow0}\frac{f\left(1+h\right)-3}{h}[/math][br][br]You might think we're done, but the problem is we still can't really "see" what's going on. If you try plugging 0 in for [code]h,[/code] in the numerator you will end up with [code]f(1)-3=3-3=0[/code]. Then 0 (numerator) divided by 0 (denominator) is undefined. [br][br]The trick is to continue re-writing equivalent algebraic statements to see if it becomes easier to understand.[br][br]For instance, probably the only thing we can do is use the actual code for the function, [code]f(x)=x^2+2x[/code], and see if that does anything to make this easier to understand. Here are those calculations:[br][br][math]\lim_{h\longrightarrow0}\frac{f\left(1+h\right)-3}{h}=\lim_{h\longrightarrow0}\frac{\left(1+h\right)^2+2\left(1+h\right)-3}{h}=\lim_{h\longrightarrow0}\frac{1+2h+h^2+2+2h-3}{h}=\lim_{h\longrightarrow0}\frac{4h+h^2}{h}[/math][br][br][math]=\lim_{h\longrightarrow0}4+h[/math][br][br]The last form of this is actually quite easy to ready as a plain English sentence: The limit of the expression [code]4+h [/code]when [code]h[/code] goes to 0. Don't overthink this. This is just 4 for obvious reasons. [br][br]That's it.[br][br]Because we've done algebraic manipulations that haven't changed the value of the [code]slope(g)[/code], the last limit and the first limit are equal. The conclusion is that [br][br][math]\lim_{h\longrightarrow0}slope\left(g\right)=4[/math][br][br]This all might look complicated and unlike anything you've ever seen, but if you understood the pattern detection in the previous activity, you understand the above calculations. This calculation is the algebraic confirmation of the pattern detection you did in the previous activity. [br][br]Side-note: In a more traditional calculus course a tremendous amount of time is spent with these types of calculations. One particular expression occurs so frequently in traditional calculus courses and texts, it gets a special name: [b]the differential quotient, [/b]and it's the algebraic bane of many a calculus student. [br][br][math]\lim_{h\longrightarrow0}\frac{f\left(x+h\right)-f\left(x\right)}{h}[/math][br][br]Students usually do oodles of calculations with this expression for different functions [code]f[/code] and different values of [code]x[/code]. We did one. We used[code] f(x)=x^2+2x[/code] and and an x value of 1.[br][br]For better or worse (I think for better), we won't do any more of these calculations. We're going to continue to rely on Geogebra to do calculations for us and built conceptual understanding by observing the computer applets, and [i]not[/i] algebraic calculations.[br][br]In the next activity we'll see a second interactive demonstration that will help you crystalize your understanding of the concept of a limit.