一下都省略极限过程x→0设 A = lim(cosx+sinx)^1/x,则lnA = lim ln(cosx+sinx)/x= lim [ln(cosx+sinx)]'/x'【L'Hospital法则】= lim (cosx+sinx)'/(cosx+sinx)= lim (-sinx+cosx)/(cosx+sinx)= lim (0+1)/(1+0)= 1于...