求log₂(x+1）-log₂(3x-1）的值域
原式=log₂[(x+1)/(3x-1)]
由(x+1)/(3x-1)=(x+1)/3(x-1/3)>0可知 x1/3.
x→±∞limlog₂[(x+1)/(3x-1)]=x→±∞limlog₂[(1+1/x)/(3-1/x)]=log₂(1/3)=-log₂3
x→1/3⁺lim log₂[(x+1)/(3x-1)]=+∞
x→ -1⁻limlog₂[(x+1)/(3x-1)=-∞
故该函数的值域为(-∞,-log₂3)∪(-log₂3,+∞)