,(sinx+cosx)/cos2x
=（sinx+cosx)/(cos²x-sin²x)
=(sinx+cosx)/[(cosx+sinx)(cosx-sinx)]
=1/(cosx-sinx)
∴x-->π/4时,1/(cosx-sinx)-->∞
即原式极限不存在
lim(x-->π/4)(sinx+cosx)/cos2x=∞lim(x-->π/4)(sinx-cosx)/cos2x=lim(x-->π/4)(sinx-cosx)/[(cosx+sinx)(cosx-sinx)]=lim(x-->π/4)1/(cosx+sinx)=√2/2