x->0
∫(0->x^(2/3) e^(t^2/2) dt
~∫(0->x^(2/3) (1+ t^2/2) dt
=[t+(1/6)t^3]|(0->x^(2/3)
=x^(2/3)+ (1/6)x^2
ln[∫(0->x^(2/3) e^(t^2/2) dt + 1 + x^(2/3) ]
~ln[1+(1/6)x^2]
~(1/6)x^2
lim(x->0) x^2/ln[∫(0->x^(2/3) e^(t^2/2) dt + 1 + x^(2/3) ]
=lim(x->0) x^2/[(1/6)x^2]
=6