Question

determine the asymptotes of the graph of f(x)= ((x^2-3x+2)/(x^2-1))

Answer 1

\[f(x)=\frac{x^2-3x+2}{x^2-1}=\frac{(x-2)(x-1)}{(x-1)(x+1)}=\begin{cases}\dfrac{x-2}{x+1}&\text{for }x\not=\pm1\\\\\text{undefined}&\text{for }x=\pm1\end{cases}\]
The difference between the singularities at \(x=-1\) and \(x=1\) is the \(x=-1\) is nonremovable and hence generates a vertical asymptote, whereas the one at \(x=1\) is removable and produces a hole.
\[\lim_{x\to\pm\infty}\frac{x^2-3x+2}{x^2-1}=1\]
which means \(y=1\) is a horizontal asymptote.