horizontal asymptote

$\dfrac {14x}{7x^2+1} = \dfrac {\frac {14}{x}}{7 + \frac{1}{x^2}} \\ \mbox{As }x \to \infty \mbox{ we have } \\ \dfrac {14}{x} \to 0 \\ \dfrac 1{x^2} \to 0 \\ \mbox{so }\dfrac {14x}{7x^2+1} =\dfrac {\frac {14}{x}}{7 + \frac{1}{x^2}} \underset{x\to \infty}{\to} \dfrac {0}{7} = 0 \\ \mbox{and thus }y=0 \mbox{ is the horizontal asymptote.}$