Let's evaluate the given limit:

$\lim_{{x \to \frac{3\pi}{2}}} \frac{\sin^2(x) + 6\sin(x) + 5}{\sin(x) + 1}$

Step 1: Analyze the behavior at $x = \frac{3\pi}{2}$

We know that: $\sin\left(\frac{3\pi}{2}\right) = -1$

Step 2: Substitute $x = \frac{3\pi}{2}$ into the expression

Substitute $\sin(x) = -1$ into the numerator and denominator:

Numerator: $\sin^2\left(\frac{3\pi}{2}\right) + 6\sin\left(\frac{3\pi}{2}\right) + 5 = (-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0$

Denominator: $\sin\left(\frac{3\pi}{2}\right) + 1 = -1 + 1 = 0$

Both the numerator and denominator equal 0, indicating an indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's Rule.

Step 3: Apply L'Hopital's Rule

Differentiate the numerator and denominator with respect to $x$:

• The derivative of the numerator $\sin^2(x) + 6\sin(x) + 5$ is: $2\sin(x)\cos(x) + 6\cos(x)$

• The derivative of the denominator $\sin(x) + 1$ is: $\cos(x)$

Step 4: Simplify the limit after differentiation

The limit now becomes: $\lim_{{x \to \frac{3\pi}{2}}} \frac{2\sin(x)\cos(x) + 6\cos(x)}{\cos(x)}$

Simplify the expression: $\lim_{{x \to \frac{3\pi}{2}}} \left( 2\sin(x) + 6 \right)$

Step 5: Substitute $x = \frac{3\pi}{2}$ again

Substitute $\sin\left(\frac{3\pi}{2}\right) = -1$:

$2(-1) + 6 = -2 + 6 = 4$

Final Answer:

The value of the limit is: $\boxed{4}$

Would you like more details on this solution or further clarification?

Here are some related questions:

1. How do we determine if L'Hopital's Rule is applicable?
2. What are other methods to solve limits involving trigonometric functions?
3. How would you handle a limit approaching $x = \pi/2$ instead?
4. Can this limit be solved without L'Hopital's Rule?
5. How does the derivative of trigonometric identities work in limit problems?

Tip: Always verify if the limit results in an indeterminate form before deciding to use L'Hopital's Rule.