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

1 Answer
May 27, 2018

Answer:

#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 !