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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 \(x \to \infty\).

Calculate the limits as x approaches infinity for the given expressions.

Learn how to solve limits as x approaches infinity for rational functions with detailed step-by-step solutions. This includes dividing by the highest power of x, analyzing asymptotic behavior, and understanding the properties of limits. Suitable for high school and early college students.

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