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lim (x,y)->(0,0) xy/[根号下(xy+1)]-1的值为
人气:437 ℃ 时间:2020-03-28 13:20:33
解答
(x,y) ->(0,0) => u = xy ->0
lim (x,y) ->(0,0) xy / [√(xy+1) - 1]
= lim u ->0 u / [√(u+1) - 1]
= lim u ->0 u *[√(u+1) + 1] / u
= lim u ->0 [√(u+1) + 1]
= 2
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