## Concrete Example

Let's begin with a concrete example. Consider the function f (t) = t2, whose graph is a parabola (Figure 2.1). Our problem is to calculate the slope of the line through the two points (1, 1) and (1 + h, 1 + 2h + h2). Let's agree to call this line €h. Both points lie on the graph of f(t) for any real number h, so the line €h always crosses the graph in two points unless h = 0, in which case it crosses in one m = t2

Figure 2.1 The parabola y = t2, with the two points [(1,1) and (1 + h, 1 + 2h + h2)] labeled.

point. Recall that slope is defined for a line y = mx + b to be the number m, and that given two points (x1; yand (x2, y2) on this line, we can calculate m by the formula

To get the slope of €h, all we have to do is plug in and calculate:

This is the answer, but some additional algebraic simplification will be worthwhile. Note the cancellation:

Taking h = 0 seems to make no sense: how could we calculate a unique slope for the line through only one point? Aren't there many lines through one point, all with different slopes? On the other hand, Equation 2.2 tells us unambiguously that at h = 0, the answer is m = 2.

The important point is that substituting h = 0 into Equation 2.1 gives 0/0, which looks like nonsense. To say a/b = k just means a = bk, so 0/0 = k would mean that 0 = 0k, but this is satisfied for any number k that you can imagine! It is much more instructive to look at the expression which appeared in Equation 2.2. Keep h > 0, but continue to lower h as much as possible towards zero without actually reaching zero. During the whole process, the small h's in the numerator and denominator still cancel. All the while, m is getting closer and closer to the value m = 2.

This situation occurs so often in mathematics that it has been given a special name: we say that the limit as h ^ 0 of m equals 2.

The notation for this is limh ^ 0m = 2. Geometrically, it's clear that our line th for h = 0 becomes a tangent line, which is a line that crosses the curve in exactly one point.

There's nothing particularly special about the point (1, 1) as far as the parabola is concerned. We could have just as easily calculated the slope at some general point (t, t 2) for some real number t by the same method. In Equation 2.2 we would have found t2 + 2ht + h2 -t2 2th + h2 (2t + h)h _ , m =-1—k—f-= —Z—= , =2t + h. (2.3)

We can now see that m depends on both t and h; it is a function of two variables, so perhaps m (t, h) would have been a better notation, and we will in fact use this later. As h approaches zero, m approaches 2t.