[tex]f(x)=\displaystyle\frac{x^2+x-2}{x+1}=\frac{x(x+1)-2}{x+1}=x-\frac{2}{x+1}
[/tex]
[tex]\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[/tex]
Consequently, the limit of [tex]f(x)[/tex] as x approaches infinity is [tex]x[/tex].
In other words, [tex]f(x)[/tex] 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.
