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Chapter 4 44

# Chapter 4 44 - —_—‘_—_——————77 306 59...

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Unformatted text preview: —____—‘___—__——————77 306 59. 60. 61. 62. 63. CHAPTER 4 APPLICATIONS OF DIFFERENTIATION _ as 1/5: 1 a: y—(e +213) => lny:;1n(e +m) so ‘ ‘ l w . I I z hmlnyzhmn—(e‘ﬂlghme—I—lHlim e Hlune—=1 é 1—.»00 2—100 :1: :c—roo e": + :17 z—aoo ex + 1 z—wo e“ . 1 . 11m (ex +37) M =11m elny = 61 = e. Iﬁm Z—>OO ’ a: m => 1n — xln w :> — £17 + 1 y _ x 1 1im1ny:1im:c1n ”B :hml—I‘LmIm—Jrﬂih 1/z_1/Ix+1) z—roo :t-roo It + 1 m—wo 1/3: z—->oo —1/\$2 2 _ = lim —m + a: : 11m x -1 so lim ( m ) = lim elny : 6—1 93 1 *1 m 1 :c *1 0r: lim< 1” > :lim[(\$+) ] :[lim(1+—>] =e—1 l 3 — y = (cos 3w)5/I => lny : §1n(cos3\$) => lim lny : 51im m 2 511m m = 0, IE m—>0 Lie—>0 CC z—>O ]_ so 111%(cos 3305/1 : 60 : 1. l/z2 1 y:(c0sa:) :> lnyz—mglncosz => 1 . Incosm H . Atancc H . —sec2a: 1 : : : : —— :> 3813131111?! \$133+ m2 \$133+ 23: \$1111): 2 2 lim (cosm)1/z2 : 111111L elny : e’l/2 : 1/\/—é z—>O" m—vO 2m — 3 29”“ 2m — 3 : : 1 1 y (n+5) :> lny OHM)“ 2\$+5 . 1n(2ac A 3) —1n(2x + 5) H . 2/(2m ! 3) — 2/(2a: + 5) . *8(2:c + 1)2 ' _ 1 — 11m 11:“; my 3511330 1 /(2:1: + 1) \$320 -2/(21: + 1)2 H» (23: # 3mm + 5) m 1 . 78(2+1/m)2 . <2\$—3)2 + _8 ﬂ : 8 => 1 = e 11:“; (2 # mm + 5/;c) \$3520 23: + 5 6 From the graph. it appears that lim w [1n(m —I— 5) — 1n cc] : 5. (t—’OO ? To prove this. we ﬁrst note that 1n(m+5)—1nm:ln\$+5 =ln(1+§> ——>1n1:0asac—>oo.Thus, 33 w 200 1 l 0 . _, 1n(m+5)—lnmﬂ 33.1.5—53 23209” [W + 5) r 1“ m1 - 1520 1/30 # Loo -1/332 . 1' — (:1: + 5) #m2 . 5x2 _ - — 1 : 5 ash—{EoI 25(3: + 5) 1 (cl—{go 1'2 + 5x ...
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