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) 的極限解: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] (洛必達法則)=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)]這個中間有具體步驟嗎，看不懂額~~~等價無窮小,當(dāng)x→0時ln(1+x)等價于x所以ln(1+sin(x/2)-2sin²x)等價于sin(x/2)-2sin²xlim(x→0)e^[(1/x)(sin(x/2)-2sin²x)]=lim(x→0)e^[1/2cos(x/2)-4sinxcosx] 這步我也看不懂~~~為什么??？洛必達法則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]就是分子分母分別求導(dǎo)數(shù)