We are given the function $r(x) = \frac{5x}{x - 8}$ and are asked to determine its behavior near the vertical asymptote $x = 8$. Specifically, we are looking for the one-sided limits as $x$ approaches 8 from the left ($x \to 8^-$) and from the right ($x \to 8^+$).

Step-by-step analysis:

1. For $x \to 8^-$ (approaching 8 from the left):

   • As $x$ approaches 8 from the left, $x - 8$ becomes a small negative number (since values of $x$ are slightly less than 8).
   • Therefore, the denominator $x - 8$ will approach $0^-$ (a small negative number).
   • The numerator $5x$ will approach $5(8) = 40$, which is positive.
   • So, the whole fraction approaches $\frac{40}{0^-}$, which tends to $-\infty$.

   Thus, $\lim_{x \to 8^-} r(x) = -\infty$.

2. For $x \to 8^+$ (approaching 8 from the right):

   • As $x$ approaches 8 from the right, $x - 8$ becomes a small positive number (since values of $x$ are slightly more than 8).
   • Therefore, the denominator $x - 8$ will approach $0^+$ (a small positive number).
   • The numerator $5x$ will approach $40$, which is positive.
   • So, the whole fraction approaches $\frac{40}{0^+}$, which tends to $+\infty$.

   Thus, $\lim_{x \to 8^+} r(x) = +\infty$.

Conclusion:

The correct option describes the behavior as: $\lim_{x \to 8^-} r(x) = -\infty \quad \text{and} \quad \lim_{x \to 8^+} r(x) = +\infty.$ This corresponds to the last option in the image.

Would you like a detailed explanation of why vertical asymptotes behave this way? Here are five related questions to explore this topic further:

1. How do you identify the location of vertical asymptotes in a rational function?
2. What happens to the graph of a function near its vertical asymptote?
3. Can a function have more than one vertical asymptote?
4. What is the difference between vertical and horizontal asymptotes?
5. How do you calculate limits involving infinity for more complex rational functions?

Tip: Always remember that vertical asymptotes occur where the denominator approaches zero but the numerator does not.