当x→∞时,x>0,所以有x-1<[x]≤x
则1-(x-1)>1-[x]≥1-x,即1-x≤1-[x]<2-x
当x→∞时,x>0,所以3x+4>0
则(1-x)/(3x+4)≤(1-[x])/(3x+4)<(2-x)/(3x+4)
注意到lim(x→∞)(1-x)/(3x+4)=lim(x→∞)(1/x-1)/(3+4/x)=-1/3
且lim(x→∞)(2-x)/(3x+4)=lim(x→∞)(2/x-1)/(3+4/x)=-1/3
则可利用夹逼定理：
由lim(x→∞)(1-x)/(3x+4)=lim(x→∞)(2-x)/(3x+4)=-1/3,且(1-x)/(3x+4)≤(1-[x])/(3x+4)<(2-x)/(3x+4)
得lim(x→∞)(1-[x])/(3x+4)=-1/3