Question

一到高数极限·······lim[（x-1）e^（π/2+arctan x) ] - (e^π)X x趋近于正无穷是 的

Answer 1

令 u = arctan x - π/2,x = tan(u+π/2) = - cotu,1/x = - tanu

x->正无穷时,u->0,tanu u

f(x) = (x-1) e^(π/2+arctanx) - x (e^π)

= e^π * [ (1- 1/x) e^(arctanx - π/2) - 1] / (1/x)

= e^π * [ (1+tanu) e^u - 1] / (-tanu) = - e^π * [ e^u + (e^u - 1) / tanu ]

lim(x->+∞) f(x) = - e^π * lim(u->0) [ e^u + (e^u - 1) / tanu ]

= - e^π * [ 1 + 1] = -2 e^π