高等数学一道求渐近线的题,这一步是怎么算出来的?

2024-11-26 08:42:44
推荐回答(1个)
回答1:

没有错
lim(x->+∞) { (x-1)e^[π/2+arctanx] - xe^(π) }
=lim(x->+∞) { x { e^[π/2+arctanx] - e^(π) ] -e^[π/2+arctanx] }
=-e^(π) + lim(x->+∞) { x [ e^[π/2+arctanx] - e^(π) ] }
=-e^(π) -e^(π)
=-2e^(π)
lim(x->+∞) x [ e^[π/2+arctanx] - e^(π) ]
=lim(x->+∞) { e^[π/2+arctanx] - e^(π) } / (1/x) (0/0 分子分母分别求导)
=lim(x->+∞) {1/(1+x^2)]. e^[π/2+arctanx] / (-1/x^2)
=lim(x->+∞) -{x^2/(1+x^2)]. e^[π/2+arctanx]
=lim(x->+∞) -e^[π/2+arctanx]
=-e^(π)