Question

高中数学，计算极限，，，，，

希望过程详细点，谢谢

Answer 1

x->0
e^(-x^2)= 1-x^2 +(1/2)x^4 +o(x^4)
1-x^2-e^(-x^2) =-(1/2)x^4 +o(x^4)
x(sin2x)^2 =4x^3 +o(x^3)
lim(x->0) [1-x^2-e^(-x^2)]/ [x(sin2x)^2]
=lim(x->0) -(1/2)x^4/ (4x^3)
=-1/8

追问

明人不说暗话，我看不懂，可以麻烦重新写一下吗？？谢谢

追答

x->0

分子

e^x = 1+x+(1/2)x^2 +o(x^2)

e^(-x^2)= 1-x^2 +(1/2)x^4 +o(x^4)

1-x^2-e^(-x^2) = 1-x^2 -[1-x^2 +(1/2)x^4 +o(x^4)] =-(1/2)x^4 +o(x^4)

分母

sinx =x+o(x^2)

(sin2x)^2 =(2x)^2 +o(x^2) =4x^2 +o(x^2)

x.(sin2x)^2 =x.[4x^2 +o(x^2)] =4x^3 +o(x^3)

lim(x->0) [1-x^2-e^(-x^2)]/ [x(sin2x)^2]
=lim(x->0) -(1/2)x^4/ (4x^3)
=-1/8

追问

谢谢