Question

求多元函数的极限

（x^2+y^2）e^(y-x) 其中x→+∞，y→-∞

Answer 1

解：∵lim(x->+∞,y->-∞)[(x-y)^2/e^(x-y)]
=lim(t->+∞)(t^2/e^t) (令t=x-y)
=lim(t->+∞)(2t/e^t) (∞/∞型极限，应用罗比达法则)
=lim(t->+∞)(2/e^t) (∞/∞型极限，应用罗比达法则)
=0
lim(x->+∞)(x/e^x)
=lim(x->+∞)(1/e^x) (∞/∞型极限，应用罗比达法则)
=0
lim(y->-∞)(ye^y)
=lim(y->-∞)[y/e^(-y)]
=lim(y->-∞)[-1/e^(-y)] (∞/∞型极限，应用罗比达法则)
=0
∴lim(x->+∞,y->-∞)[(x^2+y^2)e^(y-x)]
=lim(x->+∞,y->-∞)[((x-y)^2-2xy)/e^(x-y)]
=lim(x->+∞,y->-∞)[(x-y)^2/e^(x-y)-2(x/e^x)(ye^y)]
=lim(x->+∞,y->-∞)[(x-y)^2/e^(x-y)]-2*lim(x->+∞,y->-∞)(x/e^x)*lim(x->+∞,y->-∞)(ye^y)
=0-2*0*0
=0。