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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.

Evaluate the following limits: 
1. lim (x -> ∞) [-2x / (14 + 2x)] 
2. lim (x -> -∞) [(11x - 14) / (x^3 + 5x - 11)] 
3. lim (x -> ∞) [(x^2 - 2x - 3) / (6 - 11x^2)] 
4. lim (x -> ∞) [√(x^2 + 2x) / (6 - 12x)] 
5. lim (x -> -∞) [√(x^2 + 2x) / (6 - 12x)]

Learn how to evaluate limits of rational functions as x approaches infinity or negative infinity. This step-by-step solution covers key calculus concepts like dominant term approximation, simplification of expressions, and handling square roots for large x.

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