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To evaluate these limits, let's simplify each expression.

### a) \(\lim_{x \to \infty} (11 + x^{-2})\)

As \(x \to \infty\), \(x^{-2} = \frac{1}{x^2} \to 0\).

So, the limit is:
\[
11 + 0 = 11
\]

### b) \(\lim_{x \to \infty} \frac{3x^2 + 11x^5}{2x^7 + 4}\)

The highest power of \(x\) in the numerator is \(x^5\) and in the denominator is \(x^7\).
Divide every term by \(x^7\):

\[
\frac{\frac{3x^2}{x^7} + \frac{11x^5}{x^7}}{1 + \frac{4}{x^7}} = \frac{\frac{3}{x^5} + \frac{11}{x^2}}{2 + \frac{4}{x^7}}
\]

As \(x \to \infty\), these terms go to 0, yielding:
\[
\frac{0 + 0}{2 + 0} = 0
\]

### c) \(\lim_{x \to \infty} \left( \frac{5x^3}{x^3 + 2x^2 + 1} \right)\)

Factor out the highest degree term, \(x^3\):

\[
\frac{5x^3}{x^3(1 + \frac{2}{x} + \frac{1}{x^3})} = \frac{5}{1 + \frac{2}{x} + \frac{1}{x^3}}
\]

As \(x \to \infty\), \(\frac{2}{x} \to 0\) and \(\frac{1}{x^3} \to 0\), so the limit is:
\[
\frac{5}{1 + 0 + 0} = 5
\]

Therefore, the solutions are:
- a) 11
- b) 0
- c) 5

Evaluate the following limits of rational functions: a) lim(x→∞) (11 + x^(-2)), b) lim(x→∞) (3x^2 + 11x^5) / (2x^7 + 4), c) lim(x→∞) (5x^3 / (x^3 + 2x^2 + 1))

Learn how to evaluate limits of rational functions as x approaches infinity. Detailed solutions for a) lim(x→∞) (11 + x^(-2)), b) lim(x→∞) (3x^2 + 11x^5) / (2x^7 + 4), and c) lim(x→∞) (5x^3 / (x^3 + 2x^2 + 1)). Explore concepts of limits and asymptotic behavior. Suitable for Grades 11-12.

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