Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(1-x)^2/(x^2-2x+1)`?

Answer 1

`x=1` is not defined.

Explanation:

Actually, if we consider `f(x)=((1-x)²)/(x²-2x+1)`, it's easy to understand.
We can easily see that `x=1 => f(1)=0/0`. There's an undefined point. However, `lim_(x to 1^-) f(x)=lim_(x to 1^+) f(x) =lim ((1-x)²)/(x²-2x+1)=lim (x²-2x+1)/(x²-2x+1)=lim_(X to 0)((X+1)²-2(X+1)+1)/((X+1)²-2(X+1)+1)=lim_(X to 0) 3/3=1`
\0/ here's our answer !