Chapter 1=−(1 +h)1/hddh1−12h+13h2−14h3+· · ·=−(1 +h)1/h−12+23h−34h2+· · ·.Hencelimh→0g(h)=limh→0−(1 +h)1/h−12+23h−34h2+· · ·=−limh→0(1 +h)1/h·limh→0−12+23h−34h2+· · ·.From calculus we know thate= limh→0(1 +h)1/h, which giveslimh→0g(h) =−e−12=e2.So we havelimn→∞e−yn1/n=e2.15.The independent variable in this problem is the timetand the dependent variable is thetemperatureT(t) of a body. Thus, we will use the recursive formulas (2) and (3) on page 25withxreplaced bytandyreplaced byT. In the differential equation describing the Newton’sLaw of Cooling,
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