x->0
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)
e^(-t)= 1-t+(1/2)t^2 +o(t^2)
cost = 1 -(1/2)t^2 +o(t^2)
e^(-t).cost
=[1-t+(1/2)t^2].[1-(1/2)t^2)] +o(t^2)
=[1-(1/2)t^2)] -t[1-(1/2)t^2)] +(1/2)t^2.[1-(1/2)t^2)] +o(t^2)
= 1-t +o(t^2)
∫(0->x) e^(-t).cost dt
~∫(0->x) ( 1-t) dt
= x - (1/2)x^2
(e^x-1).∫(0->x) e^(-t).cost dt
~[x+(1/2)x^2].[x - (1/2)x^2]
~x^2
(e^x-1).∫(0->x) e^(-t).cost dt -[ln(1+x)]^2 ~ x^3
lim(x->0) { ∫(0->x) e^(-t).cost dt/[ln(1+x)]^2 - 1/(e^x -1) ]
=lim(x->0) { (e^x-1)∫(0->x) e^(-t).cost dt - [ln(1+x)]^2 }/[(e^x -1).ln(1+x)]^2]
=lim(x->0) { (e^x-1)∫(0->x) e^(-t).cost dt - [ln(1+x)]^2 }/x^3
=lim(x->0) x^3/x^3
=1