Question

一道高数题？

请大佬指教

Answer 1

直接使用洛必达法则

追答

望采纳…

Answer 2

x--->0时
(arcsinx-sinx)/(arctanx-tanx)
--->[1/√(1-x^2)-cosx]/[1/(1+x^2)-(secx)^2]
--->[(-1/2)(1-x^2)^(-3/2)*(-2x)+sinx]/[-2x/(1+x^2)^2-2sinx/(cosx)^3]
--->(x+x)/(-2x-2x)
-->-1/2.

Answer 3

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

Answer 4

根据泰勒公式：
arcsinx=x+(1/6)*x^3+o(x^3)
sinx=x-(1/6)*x^3+o(x^3)
arctanx=x-(1/3)*x^3+o(x^3)
tanx=x+(1/3)*x^3+o(x^3)
原式=lim(x->0) [x+(1/6)*x^3+o(x^3)-x+(1/6)*x^3+o(x^3)]/[x-(1/3)*x^3+o(x^3)-x-(1/3)*x^3+o(x^3)]
=lim(x->0) [(1/3)*x^3+o(x^3)]/[(-2/3)*x^3+o(x^3)]
=lim(x->0) [1/3+o(1)]/[-2/3+o(1)]
=(1/3+0)/(-2/3+0)
=-1/2