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

$x = 1$ is not defined.
Actually, if we consider f(x)=((1-x)²)/(x²-2x+1), it's easy to understand.
We can easily see that $x = 1 \implies f \left(1\right) = \frac{0}{0}$. 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