lim(x→0)[xln(1-x^2)+6(x-sinx)]⼀x^5

2025-06-24 22:25:02
推荐回答(3个)
回答1:

x->0
ln(1+x) ~ x - (1/2)x^2
ln(1-x^2) ~ -x^2 - (1/2)x^4
xln(1-x^2) ~-x^3 - (1/2)x^5
--------
sinx ~ x - (1/6)x^3 + (1/120)x^5
x-sinx ~ (1/6)x^3 - (1/120)x^5
6(x-sinx) ~ x^3 - (1/20)x^5
----------
xln(1-x^2) +6(x-sinx)
~ -x^3 - (1/2)x^5 +x^3 - (1/20)x^5
~ -(11/20)x^5
---------
lim(x->0) [xln(1-x^2) +6(x-sinx)]/x^5
=lim(x->0) -(11/20)x^5/x^5
=-11/20

回答2:

ln(1+x)~x,哪来的x-(1/2)x^2

回答3:

我也是东大的,选我最佳吧(手动滑稽)2333