Schaum's Outline of Theory and Problems of Beginning Calculus

270 THE NATURAL LOGARITHM [CHAP. 34Proof: The maximum of l/t on [l, 2) is 1, and the minimum is 4. Hence, by Problem 30.3(b),6'i(2 - 1) < (l/t) dt < l(2 - 1); that is, 4 < In 2 < 1. The strict inequalities follow from Problem 30.11.A more intuitive proof would use the area interpretation of6'(l/t) dt.We shall see later that In 2 is 0.693.. . , and we shall assume this value in what follows.PROPERTY 12.lim In x = +W.X-++a,Proof: By Property 9, we need only show that In x eventually exceeds any given positive integer k. Forx > 22k,In x > In 22k = 2k In 2 [by Property 8)so In x > 2k(l) = k [by Property 11)PROPERTY 13. lim In x = -00.x+o+Proof: Let1U = -. As x +O+, U + +W. So,X1lim In x = lim In - = lim -In U [by Property 71x+o+ u-r+(o U u-++a3=- lim lnu= -W [by Property 121U-.+CUSolved Problems34.1 Sketch the graph of y = In x.We know that In x is increasing (Property 9), that In 1 = 0 (Property l), and that 4 < In 2 < 1(Property 11).From the value y = In 2 = 0.693 . . . we can estimate the y-values at x = 4, 8, 16, . . . and atx = +,a, 6, .. . by Property 8,In 4 = 2 In 21 1 1ln-= -In2 ln-= -2ln2 ln-= -3ln2...2 4 8In 8 = 3 In 2 In 16 = 4 In 2 ...D:(ln x) = D,(x-') = - x-~ = - 1/x2 < 0 and, therefore, the graph is concave downward. There is no horizontalasymptote (by Property 12), but the negative y-axis is a vertical asymptote (by Property 13). Thegraph is sketched in Fig. 34-2. Notice that In x ussumes all real numbers us oalues.34.2 Prove: In uu = In U + In U.Inlnu=L "1 ;dtmake the change of variable w = ut (U fixed). Then dw = U dt, and the limits of integration, t = 1 and t = o,go over into w = U and w = uu, respectively,yu U 1In = -- dw = ["wudw =1 dtt