Let's go through each limit problem step by step. The goal is to evaluate these limits as $x \to \infty$ or $x \to -\infty$.

1. $\lim_{x \to \infty} \frac{-2x}{14 + 2x}$

For large $x$, the highest degree term in both the numerator and denominator dominates. We can divide the numerator and denominator by $x$:

$\lim_{x \to \infty} \frac{-2x}{14 + 2x} = \lim_{x \to \infty} \frac{-2}{\frac{14}{x} + 2}$

As $x \to \infty$, $\frac{14}{x} \to 0$, so we get:

$\lim_{x \to \infty} \frac{-2}{0 + 2} = \frac{-2}{2} = -1$

2. $\lim_{x \to -\infty} \frac{11x - 14}{x^3 + 5x - 11}$

Here, we notice that the denominator grows much faster than the numerator since the highest degree term in the denominator is $x^3$. For large negative $x$, the cubic term in the denominator will dominate. This results in:

$\lim_{x \to -\infty} \frac{11x - 14}{x^3 + 5x - 11} \approx \frac{11x}{x^3} = \frac{11}{x^2}$

As $x \to -\infty$, $\frac{11}{x^2} \to 0$, so the limit is:

$0$

3. $\lim_{x \to \infty} \frac{x^2 - 2x - 3}{6 - 11x^2}$

For large $x$, the highest degree term in both the numerator and denominator dominates. We can divide by $x^2$:

$\lim_{x \to \infty} \frac{x^2 - 2x - 3}{6 - 11x^2} = \lim_{x \to \infty} \frac{1 - \frac{2}{x} - \frac{3}{x^2}}{\frac{6}{x^2} - 11}$

As $x \to \infty$, the terms involving $\frac{1}{x}$ and $\frac{1}{x^2}$ go to 0, so this simplifies to:

$\lim_{x \to \infty} \frac{1 + 0}{0 - 11} = \frac{1}{-11} = -\frac{1}{11}$

4. $\lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{6 - 12x}$

For large $x$, $\sqrt{x^2 + 2x}$ can be approximated as $\sqrt{x^2(1 + \frac{2}{x})} = x\sqrt{1 + \frac{2}{x}}$. As $x \to \infty$, $\sqrt{1 + \frac{2}{x}} \to 1$, so:

$\lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{6 - 12x} \approx \lim_{x \to \infty} \frac{x}{-12x} = \frac{1}{-12} = -\frac{1}{12}$

5. $\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2x}}{6 - 12x}$

Similar to the previous limit, but now as $x \to -\infty$, we have to consider that $\sqrt{x^2} = -x$ for negative $x$. Therefore:

$\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2x}}{6 - 12x} \approx \lim_{x \to -\infty} \frac{-x}{-12x} = \frac{-1}{-12} = \frac{1}{12}$

Summary of Limits:

1. $-1$
2. $0$
3. $-\frac{1}{11}$
4. $-\frac{1}{12}$
5. $\frac{1}{12}$

Do you have any questions or need further clarification on any of these steps?

Here are some additional problems related to limits:

1. What is the limit $\lim_{x \to \infty} \frac{3x^2 - 4x}{x^2 + 5x + 6}$?
2. How do you evaluate $\lim_{x \to -\infty} \frac{7x^3 + 4x}{x^4 - 9}$?
3. What is the behavior of rational functions at vertical asymptotes?
4. Evaluate $\lim_{x \to 0} \frac{1}{x}$ and explain why the limit doesn't exist.
5. Explore the limit $\lim_{x \to \infty} e^{-x}$ and discuss its real-world applications.

Tip: Always check the highest degree terms in both the numerator and denominator for limits at infinity, as they often dominate the behavior of the function.