南开大学2000年数学分析考研试题.
??x?y?sin?xy?,?x,y???0,0??22x?y1. 设f?x,y???, ?0,?x,y???0,0??证明f?x,y?在点?0,0?处连续,但不可微.
2. 设f?u?具有连续的导函数,且limf??u??A?0,,
u???D???x,y?:x2?y2?R2,x,y?0?,?R?0?, (1)证明 limf?u????;
u???(2)求IR???f??x2?y2?dxdy;
Dd(3)求limIR.
R???R23.(1)叙述f?x?于区间I上一致连续的定义; (2)设f?x?,g?x?都于区间I上一致连续且有界, 证明F?x??f?x?g?x?也于I上一致连续,
4.设函数列?fn?x??于区间I上一致收敛于f?x?,且存在数列?an?,使得当x?I时,总有fn?x??an,证明f?x?于I上有界. 5.设an?0,?n?1,2,L?,Sn??ak,
k?1?an证明(1)若?收敛,则?an也收敛.
n?1n?1Snn??an(2)如果??1,??收敛,问?an是否也收敛?说明理由.
n?1n?1Sn?6.设f?x,t?于?a,?????c,d?上连续,?证明?
??a??af?x,t?dx于?c,d?上一致收敛,
f?x,d?dx收敛.
1
南开大学2000年数学分析考研试题解答
1.解:f?0,0??0,
f?x,y??x?y?xy 22x?yx?y? ?lim12x?y2??12??x?y?, x2?y22?x,y???0,0?f?x,y??f?0,0??0,于是f?x,y?在点?0,0?处连续.
显然fx?0,0??0,fy?0,0??0, 当??x????y?22?0时,
f??x,?y????fx?0,0??x?fy?0,0??y????x????y?在,
22???x??y?sin??x??y????x?2???y?2???x?2???y?2??的极限不存
所以f?x,y?在点?0,0?处不可微. 2.(1)证明 由limf??u??A?0,
u???存在M?0,当u?M时,有f??u??A, 2f?u??f?u??f?M??f?M? ?f?????u?M??f?M? ?A?u?M??f?M?, 2u???由此,可知limf?u????; (2)解 IR???f??x2?y2?dxdy
D ??2d??f??r2?rdr?00?R?1???f?R2??f?0??; ?22f?R2??f?0?IR?(3)解 lim2?lim
R???R4R???R2 2
??f??R2??2R4Rlim???2R
??4Rlim???f??R2???4A.
3、简略。
4.证明 由于?fn?x??在I上一致收敛于f?x?, 对??1,存在正整数N,当n?N时,有
fn?x??f?x??1,?x?I?,
fN?x??f?x??1,?x?I?,
f?x??f?x??fN?x??fN?x??1?aN,?x?I?, 即知f?x?在I上有界.
5、设an?0,Sn?a1?a2???an,
?1?证明: (1)当?时, ?anS?收敛;
n?1n(2) 当??1?,且nlim??S???时, ?ann?发散。 n?1Sn?(3) 当??1?,且?aan收敛时,?n1n?1S?收敛。
n?n
证明 对任意正整数n, an?Sn?Sn?1,(S0?0),
因为an?0,所以Sn?1?Sn,
(1)当??1时,利用不等式anSn?Sn?1Sn1S??S??n?Sdxn?1x?,
nN得 ?aNnSn1SN1??1N??dxn?2S,{n?n?2?Sxdx?n?1??S1x?dx??1x??an?}n?2S有界,n??ann?1S?收敛; n
3
故