The given limit is:

$\lim_{x \to \frac{3}{6}^+} \left( \frac{-21x}{3 - 6x} \right)$

Let's rewrite $\frac{3}{6}$ as $\frac{1}{2}$ for simplicity:

$\lim_{x \to \frac{1}{2}^+} \left( \frac{-21x}{3 - 6x} \right)$

Steps to solve:

1. Substitute $x = \frac{1}{2}$ into the expression:

   When we directly substitute $x = \frac{1}{2}$ into the denominator: $3 - 6 \left( \frac{1}{2} \right) = 3 - 3 = 0$ The denominator becomes zero, but the numerator: $-21 \left( \frac{1}{2} \right) = -10.5$ is non-zero. So, we have a form of $\frac{-10.5}{0}$, which is an indication of either a positive or negative infinity, depending on the behavior of the function near $x = \frac{1}{2}^+$ (from the right-hand side).

2. Analyze the sign of the denominator as $x$ approaches $\frac{1}{2}^+$:

   For $x$ values slightly greater than $\frac{1}{2}$, let's examine the denominator: $3 - 6x$ As $x \to \frac{1}{2}^+$, $6x$ becomes slightly greater than 3, so the denominator $3 - 6x$ will be slightly negative.

3. Conclusion on the limit:

   • The numerator is always negative ($-21x$) as $x > 0$.
   • The denominator approaches zero from the negative side as $x \to \frac{1}{2}^+$.

   Therefore, the expression approaches positive infinity since we have a negative number divided by a negative number (which results in a positive large value).

Thus, the limit is:

$\lim_{x \to \frac{1}{2}^+} \left( \frac{-21x}{3 - 6x} \right) = +\infty$

Would you like further explanation, or have any questions?

Related Questions:

1. What happens if we approach the limit from the left, $x \to \frac{1}{2}^-$?
2. How do you evaluate limits involving infinity more generally?
3. What is the behavior of rational functions near vertical asymptotes?
4. How would you apply L'Hôpital's Rule to limits involving $\frac{0}{0}$ or $\frac{\infty}{\infty}$?
5. Can you explain the concept of one-sided limits in more detail?

Tip: When analyzing limits with zero in the denominator, always check whether the sign of the denominator changes as $x$ approaches from different sides.