Question

求多元函数的极限 lim(x^2+y^2)^(x^2*y^2) x->0,y->0

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.