Precalculus (6th Edition) Blitzer

Write $h\left( x \right)$ in the form $\frac{p\left( x \right)}{q\left( x \right)}$. \begin{align} & h\left( x \right)=\frac{15{{x}^{3}}}{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 3. \begin{align} & h\left( x \right)=\frac{15{{x}^{3}}}{3{{x}^{2}}+1} \\ & =\frac{\frac{15{{x}^{3}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}} \end{align} Now, when x tends to infinity, $\frac{1}{x}$ tends to zero. \begin{align} & g\left( x \right)=\frac{15{{x}^{3}}}{3{{x}^{2}}+1} \\ & =\frac{\frac{15{{x}^{3}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{3}}}+\frac{1}{{{x}^{3}}}} \\ & =\frac{15x}{\frac{3}{x}+\frac{1}{{{x}^{3}}}} \end{align} It can be easily observed that, the highest degree of x is in the numerator. Due to this, when x tends to infinity, the numerator also tends to infinity. Therefore, there is no horizontal asymptote.