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求极限lim(3-根号下x^2+y^2+9)/(x^2+y^2),x→0,y→0
人气:233 ℃ 时间:2020-01-31 18:07:21
解答
lim(x→0,y→0)[3-√(x^2+y^2+9]/(x^2+y^2)=lim(x→0,y→0)[9-((x^2+y^2+9)]/(x^2+y^2)[3+√(x^2+y^2+9)]=lim(x→0,y→0)(-1)/[3+√(x^2+y^2+9)]=-1/6
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