\(f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1}\)
\(\displaystyle\lim_{x\to\infty}\left\{f(x)-x \right\}=\lim_{x\to\infty}\frac{2}{x+1}=\lim_{x\to\infty}\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}
=0\)
Consequently, the limit of \(f(x)\) as x approaches infinity is \(x\).
In other words, \(f(x)\) approaches the line y=x,
so oblique asymptote is y=x.
I'm Japanese, if you find some mistakes in my English, please let me know.
