Question

find the equation of the horizontal asymptote of the function $f(x) = \frac{x^3 + 1}{5x^2 - 3x^3}$.
(1 point)
$\circ$ $y = \frac{5}{3}$
$\circ$ $y = -\frac{1}{3}$
$\circ$ $y = 0$
$\circ$ $y = \frac{1}{5}$

Explanation:

Step1: Identify highest degree term

The highest power of $x$ in numerator and denominator is $x^3$.

Step2: Divide all terms by $x^3$

$$\begin{align*} f(x)&=\frac{\frac{x^3}{x^3}+\frac{1}{x^3}}{\frac{5x^2}{x^2}-\frac{3x^3}{x^3}}\\ &=\frac{1+\frac{1}{x^3}}{\frac{5}{x}-3} \end{align*}$$

Step3: Evaluate limit as $x\to\pm\infty$

As $x\to\pm\infty$, $\frac{1}{x^3}\to0$ and $\frac{5}{x}\to0$.

$$ \lim_{x\to\pm\infty}f(x)=\frac{1+0}{0-3}=-\frac{1}{3} $$

Step4: State horizontal asymptote

The horizontal asymptote is $y=-\frac{1}{3}$.

Answer:

$\boldsymbol{y=-\frac{1}{3}}$