微积分习题课一(多元函数极限、连续、可微及偏导)题目-777705511
习题课(多元函数极限、连续、可微及偏导)
一.累次极限与重极限 例.1 f?x,y?=例.2 例.3
11?xsin?ysin,x?y?0?yx??0,x?y?0?
?3xy?f(x,y)??x2?y2??0x2?y2?0x2?y2?0
y?0x?0x?0y?0x2y2f(x,y)?22xy?(x?y)2,证明:limlimf?x,y??limlimf?x,y??0,
而二重极限limf?x,y?不存在。
x?0y?0
一般结论:
重极限与累次极限没有关系
重极限(x,y)?(x0,y0)limf(x,y)与累次极限limlim均存在,则有 x?x0y?y0f(x,y),y?y0x?x0limlimf(x,y)(x,y)?(x0,y0)limf(x,y)=limlimx?x0y?y0f(x,y)?limlimf(x,y)y?y0x?x0
x?x0y?y0limlimf(x,y),y?y0x?x0limlimf(x,y)均存在但不等,存在 (x,y)?(x0,y0)limf(x,y)不
二.多元函数的极限与连续,连续函数性质
例.4 求下列极限: (1)
(x,y)?(0,0)(x,y)?(1,0)x?y?1x?y?1lim(x?y); (2)
lim(x?y)ln(x2?y2);
; (4)limxx??y??(3)
sin(xy)(x,y)?(0,0)xlimx?y2?xy?y2;
(5)lim(xx???y???2?y2)e?(x?y)。
例.5 证明:极限
(x,y)?(?,?)lim(xyx2?y)x?022.
例.6 若z?f?x,y?在R上连续, 且
22x2?y2???limf?x,y????, 证
明 函数f在R上一定有最小值点。
例.7 f(x)在R上连续,且 (1) x?0时, f(x)?0 (2) ?c?0, f(cx)?cf(x)
n例.8 若
f(x,y)在(0,0)点的某个邻域内有定义,
f(x,y)?x2?y2x?y22f(0,0)?0,且
(x,y)?(0,0)lim?a
微积分习题课一(多元函数极限、连续、可微及偏导)题目-777705511
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