lim (x→0)[∫上x下0 cos(t²)dt]/x=lim(x→0)cos(x²) 0/0型,用洛比达法则 =1lim( x→0)[∫上x下0 ln(1+t)dt]/(xsinx) =lim(x→0)ln(1+x)/(sinx+xcosx) 0/0型,用洛比达法则=lim(x→0) [1/(1+x)]/(cosx+cosx...