We are given the limit:

$\lim_{{y \to \pi}} \frac{{\sin^2 y}}{{1 + \cos y}}$

Step-by-Step Solution:

1. Evaluate the expression at $y = \pi$:

   • $\sin(\pi) = 0$
   • $\cos(\pi) = -1$

   Substituting these values into the expression:

   $\frac{{\sin^2(\pi)}}{{1 + \cos(\pi)}} = \frac{{0^2}}{{1 + (-1)}} = \frac{0}{0}$

   This gives an indeterminate form $\frac{0}{0}$, which means we can apply L'Hôpital's Rule.

2. Apply L'Hôpital's Rule:

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

   • Derivative of $\sin^2 y$ using the chain rule: $\frac{d}{dy} \left( \sin^2 y \right) = 2\sin y \cdot \cos y$
   • Derivative of $1 + \cos y$: $\frac{d}{dy} \left( 1 + \cos y \right) = -\sin y$

   Now the limit becomes: $\lim_{{y \to \pi}} \frac{{2\sin y \cos y}}{{-\sin y}}$

3. Simplify the expression:

   We can cancel out $\sin y$ in the numerator and denominator (as $y \to \pi$, $\sin y \neq 0$):

   $\lim_{{y \to \pi}} \frac{{2 \cos y}}{{-1}} = -2 \cos(\pi)$

4. Substitute $\cos(\pi) = -1$:

   $-2 \times (-1) = 2$

Final Answer:

$\lim_{{y \to \pi}} \frac{{\sin^2 y}}{{1 + \cos y}} = 2$

Would you like more details or further clarification?

Here are 5 related questions for practice:

1. What is the limit $\lim_{{x \to 0}} \frac{{\sin x}}{{x}}$?
2. How do you apply L'Hôpital's Rule to $\lim_{{x \to 0}} \frac{{\tan x}}{{x}}$?
3. What happens to $\lim_{{x \to 0}} \frac{{x^2}}{{1 - \cos x}}$?
4. Evaluate $\lim_{{x \to 0}} \frac{{1 - \cos x}}{{x^2}}$.
5. How does the squeeze theorem help solve limits involving trigonometric functions?

Tip: When you encounter an indeterminate form such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$, try L'Hôpital's Rule by differentiating the numerator and denominator separately.