lim(x→∞)[sin(2/x)+cos(1/x)]^x的極限.需要詳細(xì)步驟.
lim(x→∞)[sin(2/x)+cos(1/x)]^x
=lim(x→∞)[1+（sin(2/x)+cos(1/x)-1）]^x
=lim(x→∞)[1+（sin(2/x)+cos(1/x)-1）]^[1/（（sin(2/x)+cos(1/x)-1））]*[x*（（sin(2/x)+cos(1/x)-1））]
=e^lim[x*（（sin(2/x)+cos(1/x)-1））]
=e^lim(t→0)[（（sin(2t)+cos(t)-1））/t]
=e^lim(t→0)[2（sin(2t)/2t+(cos(t)-1）/t)]
=e^(2-0)=e^2