Answer:
[tex]\boxed{y = x-5}.[/tex]
Step-by-step explanation:
If the oblique asymptote exists, then its equation is:
[tex]y=mx+b,[/tex]
where
[tex]m = \lim\limits_{x\to\pm\infty}\dfrac{g(x)}{x}[/tex]
and
[tex]b = \lim\limits_{x\to\pm\infty}(g(x)-mx).[/tex]
Computing the limits with L'Hôpital's rule, we get:
[tex]m = \lim\limits_{x\to\pm\infty} \dfrac{x^2-3x-5}{x(x+2)}=\dfrac{x^2-3x-5}{x^2+2x}\overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\pm\infty} \dfrac{2x-3}{2x+2} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\pm\infty} \dfrac{2}{2} = 1.[/tex]
And:
[tex]b = \lim\limits_{x\to\pm\infty}\left(\dfrac{x^2-3x-5}{x+2}-x\right)= \lim\limits_{x\to\pm\infty}\dfrac{x^2-3x-5-x^2-2x}{x+2}=\\\\=\lim\limits_{x\to\pm\infty}\dfrac{-5x-5}{x+2} \overset{\frac{\infty}{\infty}}{=} \lim\limits_{x\to\pm\infty}\dfrac{-5}{1}= -5.[/tex]
So the oblique asymptote is given by:
[tex]\boxed{y = x-5}.[/tex]
