Question

罗比塔法则 lim(2/3.14*arctanX)的x次方，x趋向于无穷，答案是多少

Answer 1

(x->+oo)lim[(2/pi)arctanx]^x
=(x->+oo)lime^[xln[(2/pi)arctanx]
=(x->+oo)lime^{xln[1+[(2/pi)arctanx-1]}
=(x->+oo)lime^{x[(2/pi)arctanx-1]}
因为(x->+oo)x[(2/pi)arctanx-1]
=(x->+oo)[(2/pi)arctanx-1]/(1/x)
令t=1/x,则t->0+
(x->+oo)[(2/pi)arctanx-1]/(1/x)
=(t->0+)[(2/pi)arctan(1/t)-1]/t,为0/0运用罗比达法则
=(t->0+)(2/pi)(-1/t^2)/[1+1/t^2]
=(t->0+)-(2/pi)/[t^2+1]
=-2/pi
所以
=(x->+oo)lime^{x[(2/pi)arctanx-1]}
=e^(-2/pi)