原式=lim(x→0)sin(x/2)+lim(x→0)cos2x
=0+1=1不好意思是求Lim(x→0)[sin（x/2）+cos2x]^(1/x) 的极限解:原式=lim(x→0)e^[(1/x)ln(sin(x/2)+1-2sin²x)]=lim(x→0)e^[(1/x)ln(1+sin(x/2)-2sin²x)]=lim(x→0)e^[(1/x)(sin(x/2)-2sin²x)]=lim(x→0)e^[1/2cos(x/2)-4sinxcosx] (洛必达法则)=lim(x→0)e^[1/2-0]=e^(1/2)lim(x→0)e^[(1/x)ln(1+sin(x/2)-2sin²x)=]lim(x→0)e^[(1/x)(sin(x/2)-2sin²x)]这个中间有具体步骤吗，看不懂额~~~等价无穷小,当x→0时ln(1+x)等价于x所以ln(1+sin(x/2)-2sin²x)等价于sin(x/2)-2sin²xlim(x→0)e^[(1/x)(sin(x/2)-2sin²x)]=lim(x→0)e^[1/2cos(x/2)-4sinxcosx] 这步我也看不懂~~~为什么啊？洛必达法则lim(x→0)e^[(1/x)(sin(x/2)-2sin²x)]=lim(x→0)e^[(sin(x/2)-2sin²x)/x]=lim(x→0)e^[[(sin(x/2)-2sin²x)]'/x']=lim(x→0)e^[1/2cos(x/2)-4sinxcosx]就是分子分母分别求导数