lim(x->0) {∫[0,x] sintln(1+t)dt-1/3X^3+1/8x^4]} /(x-sinx)(e^x^2-1)
【首先用Taylor公式: x-sinx = x^3/3!+o(x^3) ,e^(x^2) -1= x^2+o(x^2) 】
=lim(x->0) {∫[0,x] sintln(1+t)dt-1/3X^3+1/8x^4]} /(x^3/3!+o(x^3))(x^2+o(x^2))
【等價無窮小替換：(x^3/3!+o(x^3))(x^2+o(x^2))=(x^5/6+o(x^5)) ~ x^5/6 】
= 6*lim(x->0) {∫[0,x] sintln(1+t)dt-1/3X^3+1/8x^4]} /x^5
【羅必塔法則,變動上限求導(dǎo)】
= 6*lim(x->0) {sinxln(1+x)- x^2+1/2*x^3 }/5x^4
【用Taylor公式:sinxln(1+x)=(x-x^3/3!+o(x^3))(x-x^2/2+x^3/3+o(x^3))=x^2-x^3/2+x^4/6+o(x^4)】
= 6/5*lim(x->0) {x^4/6 +o(x^4) }/x^4
=1/5