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Let's go through each limit step-by-step:

### A. \( \lim_{{x \to 0}} \frac{\sin(11x)}{6 \sin(4x)} \)

As \( x \to 0 \), both the numerator \( \sin(11x) \) and the denominator \( \sin(4x) \) approach 0, creating an indeterminate form \( \frac{0}{0} \). We can use L'Hôpital's Rule, which involves differentiating the numerator and the denominator with respect to \( x \):

1. Differentiate the numerator: \( \frac{d}{dx}(\sin(11x)) = 11 \cos(11x) \).
2. Differentiate the denominator: \( \frac{d}{dx}(6 \sin(4x)) = 6 \cdot 4 \cos(4x) = 24 \cos(4x) \).

So the limit becomes:
\[
\lim_{{x \to 0}} \frac{11 \cos(11x)}{24 \cos(4x)}.
\]

As \( x \to 0 \), \( \cos(11x) \to 1 \) and \( \cos(4x) \to 1 \), so we get:
\[
\frac{11}{24}.
\]

**Answer for A:** \( \frac{11}{24} \).

---

### B. \( \lim_{{x \to \infty}} \frac{10x^3 + 11x}{e^{6x}} \)

As \( x \to \infty \), the numerator \( 10x^3 + 11x \) grows polynomially, while \( e^{6x} \) grows exponentially. Exponential growth outpaces polynomial growth, so this limit will approach 0.

**Answer for B:** \( 0 \).

---

### C. \( \lim_{{x \to 0}} \frac{4e^{3x} - 4}{\sin(12x)} \)

As \( x \to 0 \), the numerator \( 4e^{3x} - 4 \) approaches 0, and so does the denominator \( \sin(12x) \), creating a \( \frac{0}{0} \) indeterminate form. We'll use L'Hôpital's Rule:

1. Differentiate the numerator: \( \frac{d}{dx}(4e^{3x} - 4) = 4 \cdot 3 e^{3x} = 12e^{3x} \).
2. Differentiate the denominator: \( \frac{d}{dx}(\sin(12x)) = 12 \cos(12x) \).

The limit now becomes:
\[
\lim_{{x \to 0}} \frac{12 e^{3x}}{12 \cos(12x)} = \lim_{{x \to 0}} \frac{e^{3x}}{\cos(12x)}.
\]

As \( x \to 0 \), \( e^{3x} \to 1 \) and \( \cos(12x) \to 1 \), so the limit is:
\[
1.
\]

**Answer for C:** \( 1 \).

---

### D. \( \lim_{{x \to \infty}} \frac{-5x^3 + 11x^2 + 9x}{9 + 8x + 12x^3} \)

As \( x \to \infty \), the terms of the highest power of \( x \) dominate in both the numerator and the denominator. So, we only need to consider the \( x^3 \) terms in both.

The expression simplifies to:
\[
\lim_{{x \to \infty}} \frac{-5x^3}{12x^3} = \frac{-5}{12}.
\]

**Answer for D:** \( -\frac{5}{12} \).

---

Let me know if you would like further details on any of these solutions or have additional questions!

### Follow-Up Questions

1. What are the main steps of L'Hôpital's Rule and when should it be applied?
2. Why does exponential growth dominate polynomial growth as \( x \to \infty \)?
3. How does simplification using dominant terms work in limits?
4. What are common indeterminate forms in calculus, and how do we handle them?
5. Why does \( \sin(kx) / x \to k \) as \( x \to 0 \)?

**Tip:** For limits at infinity involving polynomials and exponentials, remember that exponential functions grow faster than any polynomial.

Evaluate the following limits. Use L'Hôpital's Rule as needed. If a limit does not exist, enter DNE as your answer.

Learn how to evaluate limits involving indeterminate forms with L'Hôpital's Rule and simplify limits at infinity using dominant terms. Includes detailed step-by-step solutions for limits approaching zero and infinity, suitable for advanced high school and early college calculus students.

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