高数题(有关极限)？

Question

第7题的b c怎么做，求学霸指点

Answer 1

如图，仅供参考，希望可以帮你

Answer 2

60道典型的高数极限题｜附带解析！！！转给你身边要考高数的朋友

大学考试笔记墙
2019-12-27
60道最典型的高数极限题[good]附带解析！！！

要考高数的同学一定要收好啊，真的及格必备！这几天就考试了，请帮忙转给你身边要考高数的朋友，一定可以帮到他们

Answer 3

(7)
x->0

tanx = x+(1/3)x^3 +(2/15)x^5 +o(x^5)
arctanx=x-(1/3)x^3 +(1/5)x^5 +o(x^5)
tanx.arctanx
[x+(1/3)x^3 +(2/15)x^5 +o(x^5)].[x-(1/3)x^3 +(1/5)x^5 +o(x^5)]
=x[x-(1/3)x^3 +(1/5)x^5+o(x^5)]+(1/3)x^3.[x-(1/3)x^3 +(1/5)x^5 +o(x^5)]
+(2/15)x^5.[x-(1/3)x^3 +(1/5)x^5 +o(x^5)]
=[x^2-(1/3)x^4 +(1/5)x^6+o(x^6)]+[(1/3)x^4-(1/9)x^6 +o(x^6)]
+[(2/15)x^6+o(x^6)]
=x^2 +(1/5 -1/9 +2/15)x^6 +o(x^6)
=x^2 +[(9-5+6)/45]x^6 +o(x^6)
=x^2 +(2/9)x^6 +o(x^6)
tanx.arctanx -x^2 =(2/9)x^6 +o(x^6)
lim(x->0) (tanx.arctanx -x^2)/x^6
=lim(x->0) (2/9)x^6/x^6
=2/9
(c)
x->0
sinx= x-(1/6)x^3 +o(x^3)
x-sinx= (1/6)x^3 +o(x^3)
tanx=x+(1/3)x^3 +o(x^3)
x-tanx=-(1/3)x^3 +o(x^3)
lim(x->0) (x-sinx)/(x-tanx)
=lim(x->0) (1/6)x^3/[-(1/3)x^3]
=-1/2本回答被提问者采纳