## Precalculus (6th Edition) Blitzer

$f\left( x \right)=0\text{ or }y=0$ is the equation of the horizontal asymptote.
Write $f\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$. \begin{align} & f\left( x \right)=\frac{15x}{3{{x}^{2}}+1} \\ & =\frac{p\left( x \right)}{q\left( x \right)} \end{align} To find the horizontal asymptotes divide the numerator and denominator by the highest degree of x which is 2. \begin{align} & f\left( x \right)=\frac{15x}{3{{x}^{2}}+1} \\ & =\frac{\frac{15x}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}} \\ & =\frac{\frac{15}{x}}{3+\frac{1}{{{x}^{2}}}} \end{align} Now, when x tends to infinity, $\frac{1}{x}$ tends to zero. \begin{align} & f\left( x \right)=\frac{\frac{15}{x}}{3+\frac{1}{{{x}^{2}}}} \\ & =\frac{0}{3+0} \\ & =0 \end{align} Therefore, $f\left( x \right)=0\text{ or }y=0$ is the equation of the horizontal asymptote.