lim(x->0){[√(1+tanx)-√(1+sinx)]/x³}
=lim(x->0){[(1+tanx)-(1+sinx)]/[x³(√(1+tanx)+√(1+sinx))]}
=lim(x->0){(tanx-sinx)/[x³(√(1+tanx)+√(1+sinx))]}
=lim(x->0){[tanx(1-cosx)]/[x³(√(1+tanx)+√(1+sinx))]}
=lim(x->0){(sinx/x)(sin(x/2)/(x/2))²(1/(2cosx))/[√(1+tanx)+√(1+sinx)]}
=[1*1²*(1/2)]/[√(1+0)+√(1+0)] (应用重要极限lim(z->0)(sinz/z)=1)
=1/4.