The image shows the following mathematical expression:

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

This is a limit problem, and it can be solved by recognizing that it represents an indeterminate form as $x \to \pi$, because both the numerator and denominator approach zero at this limit.

To solve this, we can apply L'Hôpital's Rule, which is useful for resolving limits of the form $\frac{0}{0}$.

Step 1: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit gives the indeterminate form $\frac{0}{0}$, you can take the derivative of the numerator and denominator separately.

• The derivative of the numerator $\sin(x)$ is $\cos(x)$.
• The derivative of the denominator $x - \pi$ is $1$.

So, the limit becomes:

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

Step 2: Evaluate the new limit

Now we simply substitute $x = \pi$ into the new expression:

$\cos(\pi) = -1$

Thus, the value of the limit is:

$\boxed{-1}$

If you have any questions or need further clarification, feel free to ask!

Related Questions:

1. What is the general form of L'Hôpital's Rule for indeterminate forms like $\frac{0}{0}$?
2. Can we apply L'Hôpital's Rule to limits involving infinity?
3. What other limits might require L'Hôpital's Rule for evaluation?
4. How does the derivative of $\sin(x)$ lead to the solution of this limit?
5. What is the significance of the sine and cosine functions in calculus limits?

Tip: When faced with indeterminate forms like $\frac{0}{0}$, always check if L'Hôpital's Rule can simplify the problem!