Question

这道求极限的题怎么做呀？思路是什么呢？

Answer 1

如图

追问

tanx的导数不是(secx)^2吗

arctanx的导数才是1/（1+x^2）吧

Answer 2

将三角函数展成泰勒级数，根据分母可知，只需取级数的前两项，剩余项均为可去余项，接下来就很简单了。

Answer 3

这个应该运用洛必达法则上下求导就可以了，可能需要运用几次洛必达法则

Answer 4

x->0
tanx = x+(1/3)x^3+o(x^3)
tan(tanx)
=tan[x+(1/3)x^3+o(x^3)]
= (x+ (1/3)x^3) + (1/3)(x+(1/3)x^3)^3 +o(x^3)
= (x+ (1/3)x^3) + (1/3)(x^3 +o(x^3)) +o(x^3)
= x + (2/3)x^3 +o(x^3)
sinx = x-(1/6)x^3+o(x^3)
sin(sinx)
=sin[x-(1/6)x^3+o(x^3)]
= (x-(1/6)x^3) - (1/6)(x-(1/6)x^3)^3 +o(x^3)
= (x-(1/6)x^3) - (1/6)(x^3+o(x^3)) +o(x^3)
= x - (1/3)x^3 +o(x^3)
tan(tanx) -sin(sinx)
=[ x + (2/3)x^3 +o(x^3) ] -[ x - (1/3)x^3 +o(x^3)]
=x^3 +o(x^3)
lim(x->0) [ tan(tanx) -sin(sinx) ]/x^3
=lim(x->0) x^3/x^3
=1

Answer 5

希望能帮到你

追答

刚刚拍的不清楚