Question

图中例1.1画波浪线的地方，那个2A是怎么来的呢？

Answer 1

由已知x—>1时limf(x)=A,lim2x^2e^(x-1)=2，所以这部分lim[……]=2A

Answer 2

lim(x->1) f(x)=A

lim(x->1) [x-arcsin(x-1) -1]/(x-1)^3

=lim(x->1) [(x-1)-arcsin(x-1) ]/(x-1)^3

x->1 , arcsin(x-1) = (x-1) +(1/6)(x-1)^3 +o((x-1)^3)

=lim(x->1) [(x-1)-(x-1)-(1/6)(x-1)^3  ]/(x-1)^3

=-1/6

lim(x->1) [2x^2.e^(x-1).lim(x->1)f(x)]

lim(x->1)f(x) =A

=A.lim(x->1) 2x^2.e^(x-1)

代入 x=1

=A.(2)

=2A

所以

lim(x->1) [x-arcsin(x-1) -1]/(x-1)^3 + lim(x->1) [2x^2.e^(x-1).lim(x->1)f(x)]

=-1/6 +2A

=>

A= -1/6+2A

A= 1/6