得分评卷人 四、综合题（本大题共10分） 给定函数 xy≠0. 证明：该函数在(0,0)点存在两个偏导数，但是它在(0,0)点不连续 证明由偏导数的定义和极限的性质推出 色+80-i △x -巴0+4P+P-0+的 △r (△x)2 -dimto Ar dim Ar =0. 盟一点o+0 △y = 02+(0+△y)2-(02+02) -只g-只4=0 即f红，)在(0,0)点存在两个偏导数.但是 g》-g,0=四+的=四0 ge》=fe动-g1=1, 所以f(红，)在(0,0)点不存在极限，从而它在该点不连续 数学分析四试题第8页（共8页）© µò< o!nÜK £ ŒK
10 © ¤ ‰½¼ê f(x, y) =  x 2 + y 2 xy = 0, 1 xy 6= 0. y²: T¼ê3 (0, 0) :3ü‡ ê, ´§3 (0, 0) :ØëY. y² d ê½ÂÚ45ŸíÑ ∂f ∂x     (0,0) = lim ∆x→0 f(0 + ∆x, 0) − f(0, 0) ∆x = lim ∆x→0 (0 + ∆x) 2 + 02 − (02 + 02 ) ∆x = lim ∆x→0 (∆x) 2 ∆x = lim ∆x→0 ∆x = 0, ∂f ∂y     (0,0) = lim ∆y→0 f(0, 0 + ∆y) − f(0, 0) ∆y = lim ∆y→0 0 2 + (0 + ∆y) 2 − (02 + 02 ) ∆y = lim ∆y→0 (∆y) 2 ∆y = lim ∆y→0 ∆y = 0, = f(x, y) 3 (0, 0) :3ü‡ ê. ´ limx→0 y=0 f(x, y) = limx→0 f(x, 0) = limx→0 (x 2 + 02 ) = limx→0 x 2 = 0, limx→0 y=x f(x, y) = limx→0 f(x, x) = limx→0 1 = 1, ¤± f(x, y) 3 (0, 0) :Ø34, l §3T:ØëY. êÆ©Û(II)ÁK 1 8 £
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