Question

高数 这题有简单点的算法吗？

Answer 1

a=ln(x/arctanx),
b=ln(1+x)/(1-x),
When x-->0,
lim[ln(x/arctanx)]
=lim[1+(x-arctanx)/arctanx]
~(x-arctanx)/arctanx
~(x-arctanx)/x
~[(x^3)/3]/x
=(x^2)/3
In addition,
lim[ln(1+x)/(1-x)]
=lim[1+2x/(1-x)]
~2x/(1-x).
Therefore,
lima/b
=lim[ln(x/arctanx)/ln(1+x)/(1-x)]
=lim[(x-arctanx)/x]/[2x/(1-x)]
=lim[(x^2)/3]/[2x/(1-x)]
=limx/[6(1-x)]
=0

追答

a=ln(x/arctanx),
b=ln(1+x)/(1-x),
When x-->0,
lim[ln(x/arctanx)]
=lim[1+(x-arctanx)/arctanx]
~(x-arctanx)/arctanx
~(x-arctanx)/x
~[(x^3)/3]/x
=(x^2)/3
In addition,
lim[ln(1+x)/(1-x)]
=lim[1+2x/(1-x)]
~2x/(1-x).
Therefore,
lima/b
=lim[ln(x/arctanx)/ln(1+x)/(1-x)]
=lim[(x-arctanx)/x]/[2x/(1-x)]
=lim[(x^2)/3]/[2x/(1-x)]
=limx/[6(1-x)]
=0

a=ln(x/arctanx),
b=ln(1+x)/(1-x),
When x-->0,
lim[ln(x/arctanx)]
=lim[1+(x-arctanx)/arctanx]
~(x-arctanx)/arctanx
~(x-arctanx)/x
~[(x^3)/3]/x
=(x^2)/3
In addition,
lim[ln(1+x)/(1-x)]
=lim[1+2x/(1-x)]
~2x/(1-x).
Therefore,
lima/b
=lim[ln(x/arctanx)/ln(1+x)/(1-x)]
=lim[(x-arctanx)/x]/[2x/(1-x)]
=lim[(x^2)/3]/[2x/(1-x)]
=limx/[6(1-x)]
=0

Answer 2

当x->0，则：α->0，β->0
而：lim(x->0) α/β
=lim(x->0) [lnx-ln(arctanx)] /[ln(1+x)-ln(1-x)]
=lim(x->0) {1/x -1/[arctanx*(1+x^2)]}/[1/(1+x) + 1/(1-x)]
=lim(x->0) (1/2) [arctanx*(1+x^2)-x]/[x*arctanx*(1+x^2)]
=lim(x->0) (1/2) [x*(1+x^2)-x]/x^2
=lim(x->0) (1/2) x^3/x^2
=0
所以：α是比β高阶的无穷小

追问

答案是这个

追答

刚刚对分子分母按洛必达法则分别求导的时候，分子的导数漏了乘一个1/arctanx
现在已经改过了。
我的这个方法对本题来说反而不太方便，主要是分子的导数很不好求。
但在其它很多情况，这个方法还是比较管用的，特别是对选择题，因为只要简单判断谁更高阶，其实不用太仔细算出很准确的结果。