第1个回答 2020-03-27
lim(x -> ∞)(x^2-5x+1/x^2-2x+3)^(2x+1)
= lim(x -> ∞)e^ (2x+1)ln(x^2-5x+1/x^2-2x+3)
= lim(x -> ∞)e^ [(2x+1)((x^2-5x+1/x^2-2x+3) - 1)] (Taylor expansion: ln x ~ x - 1
= lim(x -> ∞)e^[(2x+1)(-3x)/(x^2)] (after ignoring lower degree terms)
= e^(-6)本回答被网友采纳