数学分析实数与数列极限-1-10

§1.12Stolz定理2008／10／16定理１，若严格增设Abbaabnnnnnn11lim,.limlim11nnnnnnnnbbaaba即.limAbannn则,,,011AbbaaANnNnnnn时，有当,11AbbaaANNNN故,1212AbbaaANNNN…….11AbbaaAnnnn证明：⑴设Ａ为有限数从而有)()()()()()(12111211NNnnnnNNnnnnNnNnbbbbbbaaaaaabbaaAA.1AbbbabaAnNnNnnnNnNnnnNnNbabbAbababbA)1)(()1)((变形得取极限夹逼可以吗？从而，.supliminflimAbabaAnnnnnn.supliminflimAbabaAnnnnnn任意性，由AbabbAAbabbAnNnNnnNnNn)1)(sup(lim)1)(inf(lim⑵,0,11111nnnnnnnnbbaabbaaA，设,严格增na.011limnnnnnaabb,0limnnnab.limnnnba⑶.,11nnnnnnbbccacA，令设,lim11nnnnnbbcc,limnnnbc.limlimnnnnnnbcba例１..21limlimanaaaaannnn，求证解：,,21nyaaaxnnn.1limlimlim11aayyxxyxnnnnnnnnnn例２.*121limNknnkkkkn解：11)1(limkkknnnn原式11)1(kknn])1()1([112111kkkkkkncnknn2121)1()1(kkkkncnk.11)1()1(2limknknkkkn例３.)1(2limaannn)1(12)1(1122limlimaanaannnnnnn2221)1(2)32()12(11limlimaaaannannnnn02)1(122limnnaa例４.1)1ln(lnlnlimlimnnnnnn0)111ln()1ln(limlimnnnnn)1)11ln(11(nnn利用定理2.limAbannn则.11limlimnnnnnnnnbbaaba即，若且设Abbaabbbbannnnnnn11321lim,0,,000:逆命题不成立,nbn12,22,2knkknkan,1limnnnba,0122limlim,211kkbbaaknknnnnn.212)1(2limlim,1211kkbbaaknknnnnn:说明作业(数学分析习题集)习题1.9Stolz定理的应用1.3)，4)；2；3.