The expression you provided is:

$\lim_{{x \to \pi}} \frac{{12x - 2\pi}}{{5 \sin 2x}}$

Step-by-step solution:

1. Substitute $x = \pi$ directly:

   • Numerator: $12\pi - 2\pi = 10\pi$
   • Denominator: $5 \sin(2\pi)$

   Since $\sin(2\pi) = 0$, the denominator becomes 0, leading to an undefined form (infinity or a division by zero).

2. Analyze the limit: Since direct substitution leads to division by zero, this suggests we may have to apply L'Hopital's Rule because the limit is of the indeterminate form $\frac{0}{0}$.

3. Differentiate numerator and denominator:

   • The derivative of the numerator $12x - 2\pi$ is $12$.
   • The derivative of the denominator $5 \sin(2x)$ is $10 \cos(2x)$ (using the chain rule).

4. Recompute the limit: $\lim_{{x \to \pi}} \frac{{12}}{{10 \cos(2x)}}$

   • At $x = \pi$, $\cos(2\pi) = 1$.

   Therefore, the limit becomes: $\frac{{12}}{{10}} = 1.2$

Final Answer:

$\lim_{{x \to \pi}} \frac{{12x - 2\pi}}{{5 \sin 2x}} = 1.2$

Would you like further details on this solution or have any specific questions?

Here are five related questions to deepen your understanding:

1. What is L'Hopital's Rule, and when can it be applied?
2. How do you differentiate trigonometric functions like $\sin(2x)$?
3. What are common indeterminate forms in limits, and how do we resolve them?
4. Why does $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$?
5. Can L'Hopital's Rule always be used, or are there exceptions?

Tip: When you encounter an indeterminate form like $\frac{0}{0}$, consider using L'Hopital's Rule, but ensure both the numerator and denominator approach zero or infinity.