Let's solve these limits step by step.

1. $\lim_{x \to \infty} \frac{-11x}{8 + 2x}$

To find the limit as $x \to \infty$, we divide both the numerator and the denominator by $x$:

$\lim_{x \to \infty} \frac{-11x}{8 + 2x} = \lim_{x \to \infty} \frac{-11}{\frac{8}{x} + 2}$

As $x \to \infty$, $\frac{8}{x} \to 0$, so the expression simplifies to:

$\lim_{x \to \infty} \frac{-11}{2} = -\frac{11}{2}$

2. $\lim_{x \to \infty} \frac{13x - 7}{x^3 + 4x - 10}$

For this limit, we divide both the numerator and the denominator by $x^3$ (the highest power of $x$ in the denominator):

$\lim_{x \to \infty} \frac{13x - 7}{x^3 + 4x - 10} = \lim_{x \to \infty} \frac{\frac{13}{x^2} - \frac{7}{x^3}}{1 + \frac{4}{x^2} - \frac{10}{x^3}}$

As $x \to \infty$, all terms with $x$ in the denominator tend to 0:

$\lim_{x \to \infty} 0 = 0$

3. $\lim_{x \to \infty} \frac{x^2 - 2x - 2}{11 - 15x^2}$

Divide both the numerator and the denominator by $x^2$ (the highest power of $x$):

$\lim_{x \to \infty} \frac{x^2 - 2x - 2}{11 - 15x^2} = \lim_{x \to \infty} \frac{1 - \frac{2}{x} - \frac{2}{x^2}}{\frac{11}{x^2} - 15}$

As $x \to \infty$, the terms involving $x$ in the denominator tend to 0, so we are left with:

$\frac{1}{-15} = -\frac{1}{15}$

4. $\lim_{x \to \infty} \frac{\sqrt{x^2 + 14x}}{6 - 3x}$

Factor out $x^2$ from inside the square root and divide both numerator and denominator by $x$:

$\lim_{x \to \infty} \frac{\sqrt{x^2(1 + \frac{14}{x})}}{6 - 3x} = \lim_{x \to \infty} \frac{x\sqrt{1 + \frac{14}{x}}}{-3x(1 - \frac{6}{3x})}$

As $x \to \infty$, $\frac{14}{x} \to 0$, so the expression simplifies to:

$\lim_{x \to \infty} \frac{x}{-3x} = -\frac{1}{3}$

5. $\lim_{x \to \infty} \frac{\sqrt{x^2 + 14x}}{6 - 3x}$

This is the same expression as the previous one, so the result is also:

$- \frac{1}{3}$

Let me know if you'd like further clarification or more detailed steps. Here are some related questions:

1. How do you determine which term dominates in a limit as $x \to \infty$?
2. Why does dividing by the highest power of $x$ simplify limits at infinity?
3. How do we handle square roots in limit problems?
4. What happens to constants as $x \to \infty$?
5. Can limits at infinity result in finite values even if the function grows without bound?

Tip: Always divide by the highest power of $x$ in limits involving polynomials to simplify the process of finding the behavior as $x \to \infty$.