Let's break down the problem shown in the image step by step.

Given:

You are asked to evaluate the limit: $\lim_{x \to 1} \frac{\sin(\pi x)}{x - 1}$

Step 1: Checking if L'Hopital's Rule can be applied

L'Hopital's Rule can be applied when a limit results in an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

1. Substitute $x = 1$ into the expression:
   • The numerator becomes $\sin(\pi \cdot 1) = \sin(\pi) = 0$.
   • The denominator is $1 - 1 = 0$.

Thus, the limit is of the indeterminate form $\frac{0}{0}$, so Yes, L'Hopital's Rule can be applied.

Step 2: Differentiating the numerator and denominator

L'Hopital's Rule states that we can take the derivative of the numerator and the derivative of the denominator, and then re-evaluate the limit.

1. Derivative of the numerator: $\frac{d}{dx} \sin(\pi x) = \pi \cos(\pi x)$
2. Derivative of the denominator: $\frac{d}{dx} (x - 1) = 1$

Step 3: Applying the limit after differentiation

Now, the limit becomes: $\lim_{x \to 1} \frac{\pi \cos(\pi x)}{1}$ Substitute $x = 1$: $\pi \cos(\pi \cdot 1) = \pi \cos(\pi) = \pi \cdot (-1) = -\pi$

Final Answer:

The limit evaluates to $-\pi$.

Summary:

1. Can L'Hopital's rule be used? Yes.
2. What is the indeterminate form? $\frac{0}{0}$.
3. Final answer: $-\pi$.

Would you like any further clarification?

Here are five related questions to explore:

1. What are the conditions for using L'Hopital's Rule?
2. How can you handle limits where L'Hopital’s Rule does not apply?
3. What other indeterminate forms exist apart from $\frac{0}{0}$?
4. How do you differentiate trigonometric functions effectively?
5. What happens if you apply L'Hopital's Rule multiple times?

Tip: Always verify whether a limit results in an indeterminate form before applying L'Hopital's Rule.