Question

# Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph.) $g(x)=\frac{x^{3}}{x^{2}-4}$

Solution

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Let's solve the exercise:

First, we will try to find $\textbf{vertical asymptotes}$ using the rule for $\textbf{Finding Vertical Asymptotes of Rational Functions}$:

\begin{align*} P(x)&=x^3 \\ Q(x)&=x^2-4 \\ \\ Q(x)&=0 \\ x^2-4&=0 \\ x^2&=4 \\ \color{#c34632}x&\color{#c34632}=\pm 2 \\ \end{align*}

We can conclude that $x=-2$ and $x=2$ are $\textbf{vertical asymptotes}$. ($P(\pm 2)\ne0$)

Now, let's try to find $\textbf{horizontal asymptotes}$. We need to find a limit either when $x \to \infty$ or $x\to -\infty$:

\begin{align*} \lim_{x\to\infty} \left(\frac{x^3}{x^2-4}\right) &=\lim_{x\to\infty} \frac{\frac{x^3}{x^2}}{\frac{x^2}{x^2}-\frac{4}{x^2}} && \text{Divide numerator and denominator by x^2.}\\ &=\lim_{x\to\infty} \frac{x}{1-\frac{4}{x^2}} && \text{Calculate.}\\ &=\frac{\infty}{1-0} \\ &\color{#c34632}=\infty && \text{Simplify.}\\ \end{align*}

We can conclude that there are no $\textbf{horizontal asymptotes}$.

Check the graph below!

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