如何判断一个函数是否可导？

Question

如题所述

Answer 1

x->0+

分子

arcsinx = x+(1/6)x^3 +o(x^3)

(arcsinx)^x - x^x

~[x+(1/6)x^3]^x -x^x

= x^x . {  [1+(1/6)x^2]^x -1  }

~[1+(1/6)x^2]^x -1

~ e^[(1/6)x^3] -1

~ (1/6)x^3

分母

ln(1+x) = x - (1/2)x^2 +o(x^2)

[ln(1+x)]^2

=[x - (1/2)x^2 +o(x^2)]^2

=x^2 - x^3 +o(x^3)

x^2- [ln(1+x)]^2 =x^3 +o(x^3)

//

lim(x->0+) [(arcsinx)^x - x^x ]/{  x^2 -[ln(1+x)]^2 }

=lim(x->0+)  (1/6)x^3/ x^3

=1/6