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  • 求多元函數(shù)的極限

    求多元函數(shù)的極限
    (x^2+y^2)e^(y-x) 其中x→+∞,y→-∞
    數(shù)學人氣:752 ℃時間:2020-10-01 02:14:58
    優(yōu)質(zhì)解答
    ∵lim(x->+∞,y->-∞)[(x-y)^2/e^(x-y)]
    =lim(t->+∞)(t^2/e^t)(令t=x-y)
    =lim(t->+∞)(2t/e^t)(∞/∞型極限,應用羅比達法則)
    =lim(t->+∞)(2/e^t)(∞/∞型極限,應用羅比達法則)
    =0
    lim(x->+∞)(x/e^x)
    =lim(x->+∞)(1/e^x)(∞/∞型極限,應用羅比達法則)
    =0
    lim(y->-∞)(ye^y)
    =lim(y->-∞)[y/e^(-y)]
    =lim(y->-∞)[-1/e^(-y)](∞/∞型極限,應用羅比達法則)
    =0
    ∴l(xiāng)im(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.
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