数学分析下册答案

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1、1421n?êØ1Ü©ê?êÚ2ÂÈ©1ÊÙê?ê§1.ý£µþ4Úe41.y²µ(1)lim(xn+yn)6limxn+limynn→∞n→∞n→∞(2)lim(xn+yn)>limxn+limynn→∞n→∞n→∞y²µ(1)limxn,limynþkê§K{xn},{yn}þk..n→∞n→∞-αk=sup{xn},βk=sup{yn}"u´§n>k§kxn+yn6αk+βkn>kn>klsup{xn+yn}6αk+βkn>klim(xn+yn)=limsup{xn+yn}6limαk+l

2、imβk=limxn+limynn→∞k→∞n>kk→∞k→∞n→∞n→∞5µelimxn,limyn+∞.~Xµlimxn=+∞§¦'Xªm>{$k¿Â§KlimynØn→∞n→∞n→∞n→∞−∞.ù{xn}Ãþ.§{xn+yn}½Ãþ.§þã'Xªw,¤á¶elimxn=−∞§¦'Xªn→∞m>{$k¿Â§KlimynØ+∞.u´{yn}þk.§llim(xn+yn)=−∞§þã'Xªn→∞n→∞w,¤á.(2)Ïxn>inf{xn},yn>inf{yn}§xn+yn>inf{xn}+in

3、f{yn}âe(.e.¥§Kinf{xn+yn}>inf{xn}+inf{yn}§linf{xn+yn}>inf{xn}+inf{yn}n>kn>kn>kKliminf{xn+yn}>liminf{xn}+inf{yn}=liminf{xn}+liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=lim(xn+yn)>limxn+limyn.n→∞n→∞n→∞2.xn>0,yn>0§y²µ(1)limxnyn6limxn·limynn→∞n→∞n→∞(2)limxnyn>l

4、imxn·limynn→∞n→∞n→∞y²µ(1)Ï06xn6sup{xn},06yn6sup{yn}§K06xnyn6sup{xn}·sup{yn}âþ(.´þ.¥§Kk06sup{xn·yn}6sup{xn}·sup{yn}l06sup{xn·yn}6sup{xn}·sup{yn}n>kn>kn>kKlimsup{xn·yn}6limsup{xn}·sup{yn}=limsup{xn}·limsup{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn6limxn·l

5、imyn.n→∞n→∞n→∞(2)Ïxn>inf{xn}>0,yn>inf{yn}>0§Kxnyn>inf{xn}·inf{yn}>0âe(.´e.¥§Kkinf{xn·yn}>inf{xn}·inf{yn}>0linf{xn·yn}>inf{xn}·inf{yn}>0n>kn>kn>kKliminf{xn·yn}>liminf{xn}·inf{yn}=liminf{xn}·liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn>limxn·limynn→∞n→

6、∞n→∞3.elimxn3§Ké?Ûê{yn}¤áµn→∞(1)lim(xn+yn)=limxn+limynn→∞n→∞n→∞143(2)lim(xn·yn)=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞y²µlimxn=αn→∞elimyn=+∞(½−∞)§K(1)w,¤á.Ïα>0§K(2)w,¤á.n→∞Ølimyn=βkên→∞Ïlimyn=β§3{yn}f{yn}§¦limyn=ββ¤kÂñf4¥ö.kkn→∞k→∞qlimxn=α§Klimxn=α§

7、lim(xn+yn)=α+β,lim(xn·yn)=αβkkkkkn→∞k→∞k→∞k→∞eyα+β{xn+yn}Âñf4¥ö(^y{)
b{xn+yn}Âñf{xn+yn}§¦limxn+yn=γ>α+βk0k0k0→∞k0k0
Klimyn=limxn+yn−limxn=γ−α>βk0→∞k0k0→∞k0k0k0→∞k0ùβ{yn}¤kÂñf4¥gñ.u´α+βÒ´{xn+yn}¤kÂñf4.Óny§α>0§α+β{xn+yn}Âñf4¥.llim(xn+yn)

8、=α+β=limxn+limynn→∞n→∞n→∞lim(xn·yn)=αβ=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞4.¦eêþ4e4µ1(1)an=(n=1,2,···)2−n+(−1)n!1n(2)an=(−1)1+(n=1,2,···)nn(−1)(3)an=(n=1,2···)nnπ(4)an=sin(n=1,2,···)5)µ(1)§küä4fêµa2k→1,a2k+1→−1(k→∞)(k=1,