Question

2. (a) For 0<x+1, we have (x - 1)/x <Ins<:- 1. (b) For je N. > 1, we have in(+1)<1/; <In(;). (c) For n,ke N, n > 1, we have In(k+)< } <In(a + and lim . = Ink (d) Use (c) to prove that ) - 1) * = In 2 (Hint: Show that

Answer 1

2.a) net tx) = xt - lux. : tx) = 1 1 1 1 1 1 1 = 2 2 2 xxl ay 1-X co xt(a) Co., it is decreasing for a 71 2.XYy towe) < fel) salomco. axt {mx. let gea) = less -(xal) g'(x) = 1 / 2 - 1 = 12 co for xxl of is decreasing for xxl - x>1x g(x) egel) ex luxa (at) co ay luncat. at Cluxent for xyli

Adding above inequalites, t h is la ) (+)+ the latest nt ut2 nt nt3 the kn kuti knt kn lu (kut) = m ( x + 1) . Forn,KEN, nyl we have, n (kt - ln (k+ k). kn ay lum in (k+ 1) L. o lim 2.- Slim (k+ k) h2 jan 27 In lim (k+ h lin I e les lim (kt kl no ex lik & lume a felnk. Hence, lim 2 T alik = nya

an JH 25 12 - + 4+ + 2nt an +----+ zut + + ---- + un -2 (+*+..) tat- - A- id $ SWISHA - - 1 - + she - 32n enfe 2 then by ces + < ln (24 h , pulling tez and I = > t (227) - ln (2+2) < t < ln (2 + 2 ) in . wit FNs

On hu Jem <?= + <- . The t) --Wint- Sling [terf 2 + 2.)-] n um vzh x lum in (2+1) < lim (I + 2 lim in (2+ 2 ) ay lu 2 a ta v Jl la lu2 . an