Question

洛必达法则求 e^(-1/x^2)/x^1000 X趋近于0

RT，快点，正在赶高数作业
答案提示设t= 1/x^2,结果为0

Answer 1

lim(x→0) e^(-1/x^2)/x^1000
=lim(x→0) e^(-1/x^2)/(x^2)^500 (t=1/x^2)
=lim(t→+∞) t^500/e^t
用500次洛必达法则得
=0

Answer 2

可设1/x^2=y,X趋近于0则y趋近于无穷，原式变为 e^(-y)/y^(-500)最后极限为0