Question

求极限lim(x→1 y→0)(x+y-1)/[根号x-根号下(1-y)]

Answer 1

∵x²y²≤(x²+y²)²/4
∴0≤(x²+y²)^(x²y²)≤(x²+y²)^[(x²+y²)²/4]
∵lim(x->0,y->0){(x²+y²)^[(x²+y²)²/4]}=lim(t->0)[t^(t²/4)]
(令t=x²+y²)
=lim(t->0)[e^(t²lnt/4)]
(应用对数性质)
=e^[lim(t->0)(t²lnt/4)]
(应用初等函数的连续性)
=e^{lim(t->0)[lnt/(4/t²)]}
=e^{lim(t->0)[(1/t)/(-8/t³)]}
(∞/∞型极限,应用罗比达法则)
=e^{lim(t->0)[t²/(-8)]}
=e^0
=1
∴由两边夹定理知,lim(x->0,y->0)[(x²+y²)^(x²y²)]=1.

Answer 2

x→1
y→0时
x
y-1
都近乎为正数
所以就是求lim(x→1
y→0)[根号x+根号下(1-y)]＝2