Question

What is the limit of `x/(x^2-x)` as x approaches `0`?

Answer 1

First notice that

`x/(x^2-x)=x/[x*(x-1)]=cancelx/[cancelx*(x-1)]=1/(x-1)`

Hence

`lim_(x->0)x/(x^2-x)=lim_(x->0) 1/(x-1) =1/(0-1)=-1`