The limit as ${\displaystyle x}$ approaches gives you the horizontal asymptote.

${\displaystyle {\begin{aligned}&\lim _{x\to -\infty }x\\&\lim _{x\to \infty }x=L\end{aligned}}}$

You can use L'Hopital rule to help you with these limits

${\displaystyle {\begin{aligned}&f(x)=\lim _{x\to \infty }{\frac {2x}{2x^{2}}}\\&f(x)={\frac {\infty }{\infty }}\end{aligned}}}$

This is an indeterminate form so you can use L'Hopital's rule.

To do this you need to take ${\displaystyle {\frac {f'(x)}{g'(x)}}}$ . From this you get

${\displaystyle \lim _{x\to \infty }{\frac {2}{4x}}}$

This gives you

${\displaystyle {\frac {2}{\infty }}}$

This is no longer in indeterminate form so, we get

${\displaystyle L=0}$