l Hǒpital s Rule and Limits of Riemann Sums (Textbook Supplement)

Transcription

1 l Hǒpital s Rule and Limits of Riemann Sums Textbook Supplement The 0/0 Indeterminate Form and l Hǒpital s Rule Some weeks back, we already encountered a fundamental 0/0 indeterminate form, namely the it sin x x 0 x. At the same time, you ll no doubt remember that the computation of this it was geometrical in nature and was crucial to the computation to the derivative of the sine function. This it is called a 0/0 indeterminate form because the its of both the numerator and denominator are 0. We ve seen many others; the Chapter 2 exam was full of such its; perhaps you ll remember the following two: 2x 2 7x + 3 x 5 1 and x 3 x 3 x 1. Note that in both cases the it of the numerator is 0 and the it of the denominator is 0. Thus, these its, too, are 0/0 indeterminate forms. While the above its can be computed using purely algebraic methods, there is an alternative and often quicker method that can be used when algebra is combined with a little differential calculus. fx In general, a 0/0 indeterminate form is a it of the form where x a gx both fx 0 and gx 0. Assume, in addition, that f and g are both x a x a differentiable and that f and g are both continuous at x a a very reasonable assumption, indeed!. Then we have fx x a gx x a fx x a gx x a fx x a x a gx x a x a f a g a f x x a x a g x. This result we summarize as by continuity of the derivatives l Hǒpital s Rule. Let f and g be functions differentiable on some interval containing x a, and assume that f and g are continuous at x a. Then 1

2 As a simple illustration, watch this: fx f x x a. x a gx x a g x 2x 2 7x + 3 x 3 x 3 4x 7 x 3 1 5, x 3 which agrees with the answer obtained algebraically. Now try your hand at the following exercises in your textbook. Page 424: 15 18, 23 24, 29, 30, for the last five problems, first compute the it of the natural logarithm of the given expression. The real reason that I wanted to introduce l Hǒpital s Rule at this early stage is to give a method though which we could compute power sums of the form j p 1 p + 2 p + + n p, where p is a nonnegative integer. As an illustration, we first consider the case in which p 0: }{{} n terms 1 + x + x x n 1 x n 1 x 1 nxn 1 n. 1 l Hǒpital s Rule Of course, nothing new was learned, apart from providing an arcane application of l Hǒpital s Rule. But watch this: 2

3 n 1 + 2x + 3x nx n 1 d dx 1 + x + x2 + + x n d x n+1 1 dx x 1 n + 1x n x 1 x n+1 1 x 1 2 l Hǒpital s Rule nx n+1 n + 1x n + 1 x 1 2 nn + 1x n nn + 1x n 1 2x 1 nn + 1x n 1 x 1 2x 1 nn + 1x n 1 nn + 1/2. 2 l Hǒpital s Rule Even here, the critical student will not be terribly impressed, as the above formula is just that of an arithmetic sum and is already familiar from Algebra II. However, where we aren t already recapitulating what is already known occurs in the next logical step. Note first that n 2 j 2 jj 1 + j jj 1 + nn + 1/2. But, 3

4 jj x + 4 3x n n 1x n 2 d x + x x n dx 2 d 2 x n+1 1 dx 2 x 1 d nx n+1 n + 1x n + 1 dx x 1 2 [nn + 1x n nn + 1x n 1 ]x 1 2 2nx n+1 n + 1x n + 1x 1 x 1 4 nn 1x n+1 2n + 1n 1x n + nn + 1x n 1 2 x 1 3 nn + 1n 1x n 2nn + 1n 1x n 1 + nn + 1n 1x n 2 3x 1 2 nn + 1n 1x n 2 x 2 2x + 1 3x 1 2 nn 2 1/3. We conclude, therefore, that Whew! j 2 nn 2 1/3 + nn + 1/2 nn + 12n + 1/6. Stern test for the serious student: Apply the above technology to compute j 3. Limits of Some Riemann Sums Whereas the basic geometric problem of differential calculus is that of finding the slope of a line tangent to the graph of a function, the basic geometrical problem of integral calculus is that of finding the area under the graph of a function between two specified values of x. Both of these processes involve iting processes: the calculation of the slope of the tangent is seen to the iting slopes of secant lines, and the area under a curve is seen to be the iting area obtained by approximating by more and thinner approximating rectangles. Your textbook authors speak of approximations by LRAM, MRAM and RRAM, and argue that these all have the same iting values as we take more and thinner rectangles. Unfortunately, however, no explicit examples were taken up; filling this void is the primary reason for my writing this particular set of notes. 4

5 Example. Consider the problem of computing the area under the graph of y 4 x 2 from x 0 to x 2. We shall use RRAM with n rectangles of uniform width 2/n. Therefore, the right-hand edge of the j th rectangle occurs where x 2j/n, forcing this rectangle to have height 4 2j/n 2 4n 2 4j 2 /n 2 and area 24n 2 4j 2 /n 3. We sum all of these rectangles as an approximation to the true area: Sum of the approximating areas 24n 2 4j 2 /n 3. Draw the relevant picture here: Using what we re already determined, we have 24n 2 4j 2 /n n 3 j nn + 12n + 1 6n3 8 4 n + 12n n2 From the above, we conclude that the true area, given by the it as n, is: Area Under The Curve 8 4 n + 12n n 3n2 3. 5

6 Exercises. Using n RRAM with n approximating rectangles of uniform width, compute the areas under the given graphs between the given endpoints. 1. y x 2 on [0, 2]. 2. y 2 x + 4 on [1, 3]. Of course, this is a rather silly question as it s nothing 3 more than the area of a trapezoid. However, I d still like for you to compute the it of the approximating RRAM. 3. y 1 2 x2 + 2 on [ 2, 2]. 4. y x 2 6x + 5 on that portion of the x-axis where y 0. Can you see how to use translation and symmetry to make this problem simpler? 6