The given limit expression is:

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

Step-by-Step Solution:

1. Substitute $x = \pi$: Direct substitution gives: $\frac{\sin(\pi)}{\pi - \pi} = \frac{0}{0}$ This is an indeterminate form, so we apply L'Hôpital's Rule.

2. Differentiate numerator and denominator: Differentiating the numerator and denominator with respect to $x$: $\text{Numerator: } \frac{d}{dx}[\sin(x)] = \cos(x)$ $\text{Denominator: } \frac{d}{dx}[\pi - x] = -1$

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

3. Evaluate the limit: Substitute $x = \pi$: $\frac{\cos(\pi)}{-1} = \frac{-1}{-1} = 1$

Final Answer:

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

Would you like further clarification or details?

Related Questions:

1. What is L'Hôpital's Rule, and when is it applicable?
2. How do we evaluate trigonometric limits involving $\sin(x)$ and $\cos(x)$?
3. Why does substituting directly yield an indeterminate form here?
4. How do derivatives help resolve indeterminate limits?
5. What are other common indeterminate forms besides $\frac{0}{0}$?

Tip:

Always check if direct substitution yields a determinate value before applying L'Hôpital's Rule. This saves unnecessary differentiation steps!