Question

为什么下列中不能用洛必达法则？

Answer 1

可以用洛必达法则
但是你的无穷小代换有问题，ln(1+x)-x，按照你的代换为x-x是错误的
无穷小不能进行减法

Answer 2

因为你代换错了。。。。不能这样子代换

追答

正常的代换

追问

谢谢！

追答

嗯嗯

Answer 3

x->0
分子
(sinx)^2 = x^2 +o(x^2)
e^x = 1+ x +(1/2)x^2 +o(x^2)
(sinx)^2 +e^x = 1+x +(3/2)^2 +o(x^2)
ln[ (sinx)^2 +e^x ]
=ln[ 1+x +(3/2)^2 +o(x^2) ]
=[x +(3/2)x^2] -(1/2)[x +(3/2)^2]^2 +o(x^2)
=[x +(3/2)x^2] -(1/2)[x^2 +o(x)^2] +o(x^2)
=x +x^2 +o(x^2)
ln[ (sinx)^2 +e^x ] -x = x^2 +o(x^2)
分母
e^(2x) = 1+2x+ 2x^2 +o(x^2)
x^2+e^(2x) = 1+2x+ 3x^2 +o(x^2)
ln[x^2+e^(2x)]
=ln[1+2x+ 3x^2]
=[2x+ 3x^2] -(1/2)[2x+ 3x^2]^2 +o(x^2)
=[2x+ 3x^2] -(1/2)[4x^2+ o(x^2)] +o(x^2)
=2x + x^2 +o(x^2)
ln[x^2+e^(2x)] -2x = x^2 +o(x^2)
lim(x->0) { ln[ (sinx)^2 +e^x ] -x }/ { ln[x^2+e^(2x)] -2x }
=lim(x->0) x^2/x^2
=1