Math 310 -hw4 solutions8.8a, 8.10, 9.1b;Wednesday, 30 Sept 20098.8 (a) Prove the following:limn→∞hpn2+ 1-ni= 0.We first observe thatpn2+ 1-n=pn2+ 1-n√n2+ 1 +n√n2+ 1 +n=(n2+ 1)-n2√n2+ 1 +n=1√n2+ 1 +n.Letε >0 and setN:=1ε. Letn∈Nand suppose thatn > N. Then we have (note√n2+ 1 +n >0):pn2+ 1-n-0 =1√n2+ 1 +n=1√n2+ 1 +n<1n< ε.Hence,√n2+ 1-n→0.8.10 Let (sn)nbe a convergent sequence and suppose that limn→∞sn> a. Prove that there exists a numberNsuchthatn > Nimpliessn> a.Setsdenote the limit of the convergent sequence (sn)n. Then by assumption,s > aand soε:=s-a >0. Bydefinition of limit there exists a numberNsuch that for alln∈N,n > Nimplies|sn-s|< ε. Letn∈Nandsuppose thatn > N, then we havea=s-ε < sn< s+ε.Hence,sn> aas desired.9.1 (b) Using the limit theorems 9.2-9.6 and 9.7, prove the following. Justify all steps.limn→∞3n+ 76n-5=12.We will need the facts, the first was proved in class (see also Example 9.7(a)) and the second is obvious,limn→∞1n= 0limn→∞c=cwherecis a constant. Observe that our sequence may be rewritten as follows:
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