Question 1067303
Since the degree of {{{2x-6))) is {{{1}}} ,
which is less than {{{2}}} , the degree of {{{x^2-x-2}}} ,
{{{lim(x->infinity, (2x-6)/(x^2-x-2))=0}}} .
so, the horizontal asymptote is {{{highlight(y=0)}}} .
{{{graph(600,300,-20,20,-5,5,6,(2x-6)/(x^2-x-2))}}}