It seems that the extraction from the image was not successful. However, I can clearly see the expression from the image. The problem is to find:

$\lim_{{x \to \pi}} \frac{{\sin x}}{{\pi - x}}$

Let's solve this limit.

Step-by-Step Solution:

This limit is in the indeterminate form $\frac{0}{0}$ as $x \to \pi$ because:

• $\sin(\pi) = 0$
• $\pi - \pi = 0$

We can apply L'Hopital's Rule, which is used for limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. According to L'Hopital's Rule:

$\lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}}$

if the limit exists. Here, we have:

• $f(x) = \sin(x)$, so $f'(x) = \cos(x)$
• $g(x) = \pi - x$, so $g'(x) = -1$

Now applying L'Hopital's Rule:

$\lim_{{x \to \pi}} \frac{{\sin(x)}}{{\pi - x}} = \lim_{{x \to \pi}} \frac{{\cos(x)}}{{-1}}$

Since $\cos(\pi) = -1$, we get:

$\lim_{{x \to \pi}} \frac{{\cos(x)}}{{-1}} = \frac{{-1}}{{-1}} = 1$

Thus, the value of the limit is:

$\boxed{1}$

Would you like more details on any part of this solution? Here are five related questions:

1. What is L'Hopital's Rule, and when can it be applied?
2. How do you differentiate trigonometric functions like $\sin(x)$ and $\cos(x)$?
3. Can this problem be solved using a series expansion of $\sin(x)$?
4. What happens if you attempt this limit without L'Hopital's Rule?
5. Are there other indeterminate forms besides $\frac{0}{0}$?

Tip:

Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L'Hopital's Rule, as it can greatly simplify evaluating limits.