lim x→0[∫上x下0 cos(t^2)dt]/x ; lim x→0[∫上x下0 ln(1+t)dt]/(xsin

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
=1
lim( 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-xsinx) 0/0型,用洛比达法则
=1/2