You have provided an expression to find the limit of $\frac{3x}{x-6}$ as $x$ approaches 6 from the positive side (notated as $x \to 6^+$). Let's analyze and solve it.

Given the expression:

$\lim_{x \to 6^+} \frac{3x}{x-6}$

As $x$ approaches 6 from the right, the denominator $x - 6$ approaches a very small positive number since $x$ is slightly greater than 6. Hence, $x - 6$ is close to zero but positive.

For the numerator, $3x$ approaches 18 as $x$ approaches 6. Therefore, the fraction $\frac{3x}{x-6}$ will approach a very large positive number as the numerator is positive and the denominator is a small positive number close to zero.

Thus, the limit is:

$\lim_{x \to 6^+} \frac{3x}{x-6} = +\infty$

This implies that as $x$ approaches 6 from the right, the value of $\frac{3x}{x-6}$ increases without bound.

Would you like more details on this solution or have any questions? Here are some related questions you might find interesting:

1. What would be the limit of $\frac{3x}{x-6}$ as $x$ approaches 6 from the left ($x \to 6^-$)?
2. How would the limit change if it were $\frac{3x}{x+6}$ as $x$ approaches 6?
3. What is the behavior of the function $\frac{3x}{x-6}$ around values other than 6?
4. Can you explain why the limit to the positive and negative side might differ?
5. How do you find limits at infinity for similar rational functions?

Tip: Always consider the sign of the denominator when approaching a certain point, as it determines whether the limit heads towards positive or negative infinity, or if it exists at all.