高等数学经典例题

. 多元例3设f(x?y,xy)?x2?y2,求f(x,y).函数解令x?y?u,xy?v，则微分x?uv法1?v,y?u1?v,及其?f(u,v)?u2(v2?1)x2(y?1(1?v)2,?f(x,y)?)1?y.应用 Word 文档

例6求证lim(x2?y21x)sinx2?y2?0y??00证(x2?y2)sin1x2?y2?0?x2?y2?sin122x2?y2?x?y???0,????,当0?(x?0)2?(y?0)2??时，(x2?y2)sin1x2?y2?0??原结论成立． . 例9求22xy故lim(x2?y2)xy?e0?1.(x,ylim)?(0,0)(x?y).(x,y)?(0,0)解(x2?y2)?exyln(x2?y2),而例10求limxy?1?1xyln(x2?y2)?xy222(x,y)?(0,0)x?y.x2?y2(x?y)ln(x?y2)解先将函数变形?2222令x2?y2?txy?1?1(x,ylim)?(0,0)(x?y)ln(x?y)limt?0tlnt1x?y?xyx?y?1xy?1?1,?limlnt罗必达法则?11t?01limt(x,ylim)?(0,0)tt?0?1?0xy?1?1?2,t2又xyx2?y2?1，从而(x,ylim)?(0,0)xyln(x2?y2)?0,故所求极限是否存在,只须考察xyx?y, Word 文档 A. . 当(x,y)沿y?x趋于(0,0)时,有xy(x,y)?(0,0)x?y?xlimxy?limx2lim?0,y??x0?0x?yx?02x2又limxyxy(x,y)?(0,0)x?y?x?lim0x?y?limx(x?x)x?0y?x2?x?0x2??1,故limxy(x,y)?(0,0)x?y不存在,从而xy?1?1(x,ylim)?(0,0)x?y不存在.Word 文档 例4 设z?x3y2?3xy3?xy?1， 偏求?2z?2z?2z?2z?3z?x2、 ?y?x、?x?y、?y2及?x3. 微解?z?x?3x2y2?3y3?y,?z?y?2x3y?9xy2?x; ?2z ?x2?6xy2,?3z2?x3?6y2,?z?y2?2x3?18xy;分 ?2z?x?y?6x2y?9y2?1,?2z22?y?x?6xy?9y?1.