3.1   LINEAR APPROXIMATIONS AND L'HÔPITAL'S RULE For what purpose do you use a scientific calculator? If you think about it, you'll discover that there are two distinctly different jobs that calculators do for you. First, they perform arithmetic operations (addition, subtraction, multiplication and division) much faster than any of us could hope to do them. It's not that you don't know how to multiply 1024 by 1673, but rather that it is time-consuming to carry out this (albeit well-understood) calculation with pencil and paper. For such problems, calculators are a tremendous convenience, which none of us would like to live without. Perhaps more significantly, we also use our calculators to compute values of transcendental functions such as sine, cosine, tangent, exponentials and logarithms. In the case of these function evaluations, the calculator is much more than a mere convenience.

If asked to calculate sin(1.2345678) without a calculator, you would probably draw a blank. Don't worry, there's nothing wrong with your background. (Also, don't worry that anyone will ever ask you to do this without a calculator.) The problem is that the sine function is not algebraic. That is, there is no formula for sin x involving only the arithmetic operations. So, how does your calculator “know” that sin(1.2345678) 0.9440056953? In short, it doesn't know this at all. Rather, the calculator has a built-in program that generates approximate values of the sine and other transcendental functions. In this section, we take a small step into the (very large) world of approximation by developing a simple approximation method. Although somewhat crude, it points the way toward more sophisticated approximation techniques to follow later in the text. Our primary intent here is to give you a taste of how you might approach the problem of approximation. Figure 3.1 Linear approximation of f (x1). Linear Approximations Suppose we wanted to find an approximation for f (x1), where f (x1) is unknown, but where f (x0) is known for some x0 “close” to x1. For instance, the value of cos(1) is unknown, but we do know that cos(/3) = exactly and /3 1.047 is “close” to 1. We could always use as an approximation to cos(1), but we can do better. Recall that the tangent line to the curve y = f (x) at x = x0 stays close to the curve near the point of tangency. Referring to Figure 3.1, notice that if x1 is “close” to x0 and we follow the tangent line at x = x0 to the point corresponding to x = x1, then the y - coordinate of that point (y1) should be “close” to the y - coordinate of the point on the curve y = f (x) [i.e., f (x1)] . Since the slope of the tangent line to y = f (x) at x = x0 is f '(x0), the equation of the tangent line to y = f (x) at x = x0 is found from (1.1) Solving equation (1.1) for y gives us y = f (x0) + f '(x0)(x - x0). (1.2) Notice that (1.2) is the equation of the tangent line to the graph of y = f (x) at x = x0. We give the linear function defined by this equation a name, as follows.

1.1      The linear (or tangent line) approximation of f (x) at x = x0 is the function L(x) = f (x0) + f '(x0)(x - x0).

Notice that the y - coordinate y1 of the point on the tangent line corresponding to x = x1 is simply found by substituting x = x1 in equation (1.2). This gives us y1 = f (x0) + f '(x0)(x1 - x0). (1.3) We define the increments x and y by x = x1 - x0 and y = f (x1) - f (x0). Using this notation, equation (1.3) gives us the approximation    f (x1) y1 = f (x0) + f ' (x0) x.    (1.4) We illustrate this in Figure 3.2. We sometimes rewrite (1.4) by subtracting f (x0) from both sides, to yield y = f (x1) - f (x0) f ' (x0) x = dy, (1.5) where dy = f '(x0) x is called the differential of y. When using this notation, we also define dx, the differential of x, by dx = x, so that by (1.5), dy = f '(x0) dx. Figure 3.2 Linear Approximation of f(x1). We can use linear approximations to produce approximate values of transcendental functions, as in the following example.

1.1    Finding a Linear Approximation   Find the linear approximation to f (x) = cos x at x0 = /3 and use it to approximate cos (1).     From Definition 1.1, the linear approximation is defined as L(x) = f (x0) + f ' (x0) (x - x0). Here, x0 = /3, f (x) = cos x and f '(x) = - sin x. So, we have In Figure 3.3, we show a graph of y = cosx and the linear approximation to cosx for x0 = /3. Notice that the linear approximation (i.e., the tangent line at x0 = /3) stays close to the graph of y = cos x only for x close to /3. In fact, for x<0 or x>, the linear approximation is obviously quite poor. This is typical of linear approximations (tangent lines). They generally stay close to the curve only nearby the point of tangency. Figure 3.3 y = cos x and the linear approximation at x0 = /3. Although we gave you the value of x0 here, observe that if you had been given the choice, one reason you might choose x0 = /3 is that /3 is the value closest to 1 at which we know the value of the cosine exactly. Finally, an estimate of cos(1) is Notice that on your calculator you would find cos(1) 0.5403, so that we have found a fairly good approximation to the desired value.

In the following example, we derive a useful approximation to sinx, valid for x close to 0. This approximation is significant in part because it is often used in applications in physics and engineering to simplify equations involving sin x.

1.2    Linear Approximation of sin x   Find the linear approximation of f (x) = sin x, for x close to 0.     Here, f '(x) = cos x, so that from Definition 1.1, we have sin x L(x) = f (0) + f '(0) (x -0) = sin 0 + cos 0(x) = x. This says that for x close to 0, sinx x. We illustrate this in Figure 3.4, where we show graphs of both y = sinx and y = x.

Pay close attention to Figure 3.4; notice that the graph of y = x stays close to the graph of y = sin x only in the vicinity of x = 0. This indicates that the approximation sin x x is valid only for x close to 0. Also note that the farther x gets from 0, the worse the approximation becomes. Take another look at Figure 3.1, to convince yourself that this is generally true. This becomes even more apparent in the following example, where we also illustrate the use of the increments x and y. Figure 3.4 y = sinx and y = x.

1.3    Linear Approximation to Some Cube Roots   Use a linear approximation to approximate , and .     Here we are approximating values of the function . So, . Of course, the closest number to any of 8.02, 8.07 or 8.15 whose cube root we know exactly is 8. Now, (1.6) From (1.5), we have (1.7) Using (1.6) and (1.7), we get while your calculator accurately returns . Similarly, we get and while your calculator returns and , respectively. Finally, notice that to approximate , 8 is not the closest number to 25.2 whose cube root we know exactly. Since 25.2 is much closer to 27 than to 8, we write f (25.2) = f (27) + y f (27) + dy = 3 + dy. In this case, and we have compared to the value of 2.931794, produced by your calculator. It is important to recognize here that the farther the value of x gets from the point of tangency, the worse the approximation tends to be. You can see this clearly in Figure 3.5, where the linear approximation gets farther away from , as x gets farther from 8. Figure 3.5 y = and the linear approximation at x0 = 8.

Our first three examples were intended to familiarize you with the technique and to give you a feel for how good (or bad) linear approximations tend to be. In the following example, there is no exact answer to compare with the approximation. Our use of the linear approximation here is referred to as linear interpolation.

1.4    Using a Linear Approximation to Perform Linear Interpolation   The price of an item affects consumer demand for that item. Suppose that based on market research, a company estimates that f (x) thousand small cameras can be sold at the price of $x, as given in the accompanying table. Estimate the number of cameras that can be sold at $7.     x 6 10 14 f (x) 84 60 32 The closest x - value to x = 7 in the table is x = 6. (In other words, this is the closest value of x at which we know the value of f (x) .) The linear approximation of f (x) at x = 6 would look like L(x) = f (6) + f '(6)(x -6). From the table, we know that f (6) = 84, but we do not know f '(6). Further, we can't compute f '(x), since we don't have a formula for f (x). The best we can do with the given data is approximate the derivative by The linear approximation is then L(x) = 84-6(x -6). Using this, we estimate that the number of cameras sold at x = 7 would be L(7) = 84 - 6 = 78 thousand. That is, we would expect to sell approximately 78 thousand cameras at a price of $7. We show a graphical interpretation of this in Figure 3.6, where the straight line is the linear approximation (in this case, the secant line joining the first two data points).

Figure 3.6 Linear interpolation. L'Hôpital's Rule We close this section by using linear approximations to suggest a simple method for computing some challenging limits. We develop a special case of an important result known as L'Hôpital's Rule, which we more thoroughly develop in section 7.6. Look back at section 2.5, where we struggled with the limit ultimately resolving it only with an intricate geometric argument. This limit has the indeterminate form (i.e., the limits of the numerator and the denominator are both 0), but there is no way to simplify the numerator or denominator to simplify the expression. More generally, we'd like to evaluate , where . We can use linear approximations to suggest a solution, as follows. If both functions are differentiable at x = c, then they are also continuous at x = c, so that and . We now have the linear approximations f (x) f (c) + f '(c)(x - c) = f '(c)(x - c) and g(x) g(c) + g'(c)(x - c) = g'(c)(x - c), where we have used the fact that f (c) = 0 and g(c) = 0. As we have seen, the approximation should improve as x approaches c, so we would expect that if the limits exist, assuming that g'(c) 0. Note that if f '(x) and g'(x) are continuous at x = c and g'(c) 0, then . We summarize this in the following result.

1.1    (L'Hôpital's Rule)   Suppose that f and g are differentiable on the interval (a, b) except possibly at some fixed point c in (a, b) and that g'(x) 0 on (a, b) except possibly at x = c. If and exists, then

Here, we prove only the special case where f, f ', g and g ' are all continuous on all of (a, b) and g'(c) 0, while leaving the more intricate general case for Appendix A. First, recall the alternative form of the definition of derivative (found in section 2.2): Working backward, we have by continuity that Further, since f and g are continuous at x = c, we have that It now follows that which is what we wanted. With this result, certain limits become quite easy to evaluate.

1.5    Revisiting an Old Limit   Evaluate .     Again, this limit has the indeterminate form and f (x) = sinx and g(x) = x are both continuous and differentiable everywhere. Finally, 10, so that all of the hypotheses of L'Hôpital's Rule are satisfied. From the graph in Figure3.7, it appears that the limit is approximately 1. We can confirm this suspicion with L'Hôpital's Rule. We have as we proved using a complicated geometric argument in section 2.5. Figure 3.7

For some limits, you must apply L'Hôpital's Rule more than once.

1.6    A Limit Requiring Two Applications of L'Hôpital's Rule   Evaluate     Again, this has the indeterminate form and it is a simple matter to verify that the hypotheses of L'Hôpital's Rule are satisfied. In this case, the graph in Figure 3.8 indicates the limit to be approximately 0.5. From L'Hôpital's Rule, we have which again has the indeterminate form . In this case, we can verify that the hypotheses of L'Hôpital's Rule are satisfied for this new limit problem. Applying this again, it then follows that Figure 3.8

1.1

1.7    An Erroneous Use of L'Hôpital's Rule   Find the mistake in the string of equalities     From the graph in Figure 3.9, we can see that the limit is approximately 0, so 2 appears to be incorrect. The first limit has the form and the functions f (x) = x2 and g(x) = ex -1 satisfy the hypotheses of L'Hôpital's Rule. There-fore, the first equality: holds. However, notice that and L'Hôpital's Rule does not apply here. The correct evaluation is then Figure 3.9