高等数学问题,2元函数的极限求lim (1 + xy)^(1/(x+y))的极限 x→0 y→0书上解法是是原式=lim e^xy/(x+y) (x,y)→(0,0)请问高手这一步是怎么得来的,我已经知道lim (1 +x) ^1/x=e
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![高等数学问题,2元函数的极限求lim (1 + xy)^(1/(x+y))的极限 x→0 y→0书上解法是是原式=lim e^xy/(x+y) (x,y)→(0,0)请问高手这一步是怎么得来的,我已经知道lim (1 +x) ^1/x=e](/uploads/image/z/5570633-65-3.jpg?t=%E9%AB%98%E7%AD%89%E6%95%B0%E5%AD%A6%E9%97%AE%E9%A2%98%2C2%E5%85%83%E5%87%BD%E6%95%B0%E7%9A%84%E6%9E%81%E9%99%90%E6%B1%82lim++%EF%BC%881+%2B+xy%EF%BC%89%5E%281%2F%28x%2By%29%29%E7%9A%84%E6%9E%81%E9%99%90++x%E2%86%920+++y%E2%86%920%E4%B9%A6%E4%B8%8A%E8%A7%A3%E6%B3%95%E6%98%AF%E6%98%AF%E5%8E%9F%E5%BC%8F%3Dlim+e%5Exy%2F%28x%2By%29+++++++++++++++++++++++%28x%2Cy%29%E2%86%92%280%2C0%29%E8%AF%B7%E9%97%AE%E9%AB%98%E6%89%8B%E8%BF%99%E4%B8%80%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E6%9D%A5%E7%9A%84%2C%E6%88%91%E5%B7%B2%E7%BB%8F%E7%9F%A5%E9%81%93lim++%EF%BC%881+%2Bx%29+%5E1%2Fx%3De)
高等数学问题,2元函数的极限求lim (1 + xy)^(1/(x+y))的极限 x→0 y→0书上解法是是原式=lim e^xy/(x+y) (x,y)→(0,0)请问高手这一步是怎么得来的,我已经知道lim (1 +x) ^1/x=e
高等数学问题,2元函数的极限
求lim (1 + xy)^(1/(x+y))的极限
x→0
y→0
书上解法是是原式=lim e^xy/(x+y)
(x,y)→(0,0)
请问高手这一步是怎么得来的,
我已经知道lim (1 +x) ^1/x=e
x→0
高等数学问题,2元函数的极限求lim (1 + xy)^(1/(x+y))的极限 x→0 y→0书上解法是是原式=lim e^xy/(x+y) (x,y)→(0,0)请问高手这一步是怎么得来的,我已经知道lim (1 +x) ^1/x=e
对于幂指函数的极限,常取对数,所以化为e^[ln(1+xy)/(x+y)],用等价无穷小ln(1+x)~x(x→0),则ln(1+xy)替换为xy,得
原式=lim e^xy/(x+y)
两边求导
因为lim (1 +x) ^1/x=e,所以可知lim (1 +xy) ^1/(xy)=e, x→0 x→0 ,y→0
故 lim (1 + xy)^(1/(x+y))= lim(1+xy)^[(1/xy)*(xy/(x+y))]=l...
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因为lim (1 +x) ^1/x=e,所以可知lim (1 +xy) ^1/(xy)=e, x→0 x→0 ,y→0
故 lim (1 + xy)^(1/(x+y))= lim(1+xy)^[(1/xy)*(xy/(x+y))]=lime^xy/(x+y)
x→0 ,y→0 (x,y)→0 (x,y)→0
收起
令y=x, lim(x,y)趋于(0,0)xy/x+y
=lim(x趋于0)x^2/(2x)=0
令y=x^2-x,lim(x,y)趋于(0,0)xy/x+y
= lim(x趋于0)
lim (1 + xy)^(1/(x+y))
=lim (1 + xy)^[1/xy*xy(1/(x+y))](将xy当成一个数来看)
=lim e^[xy/(x+y)]
{e^[xy/(x+y)]}‘=0时有极值
求导得:(x^2+y^2)/(x+y)^2=0
x→0
y→0