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Let's solve these limits one by one by analyzing the behavior as \( n \to \infty \). Each expression involves a rational function, so we’ll simplify each limit by identifying the highest power of \( n \) in both the numerator and the denominator.

### Problem (1)
\[
\lim_{n \to \infty} \frac{2n^3 - 7n^2 + 8n^5}{5n^5 + 4n^2 - 2}
\]
The highest power of \( n \) in both the numerator and denominator is \( n^5 \). Dividing every term by \( n^5 \), we get:
\[
\lim_{n \to \infty} \frac{\frac{2n^3}{n^5} - \frac{7n^2}{n^5} + \frac{8n^5}{n^5}}{\frac{5n^5}{n^5} + \frac{4n^2}{n^5} - \frac{2}{n^5}} = \lim_{n \to \infty} \frac{0 - 0 + 8}{5 + 0 - 0} = \frac{8}{5}.
\]

### Problem (2)
\[
\lim_{n \to \infty} \frac{2n^2 - 7}{\sqrt{5n^4 + 14n^2}}
\]
The denominator can be rewritten as \( \sqrt{5n^4 + 14n^2} = n^2 \sqrt{5 + \frac{14}{n^2}} \). Dividing the numerator and denominator by \( n^2 \), we obtain:
\[
\lim_{n \to \infty} \frac{\frac{2n^2}{n^2} - \frac{7}{n^2}}{\sqrt{5 + \frac{14}{n^2}}} = \frac{2 - 0}{\sqrt{5 + 0}} = \frac{2}{\sqrt{5}}.
\]

### Problem (3)
\[
\lim_{n \to \infty} \frac{2n^2 - 7n}{-5n + 14}
\]
Dividing both the numerator and the denominator by \( n \), we get:
\[
\lim_{n \to \infty} \frac{\frac{2n^2}{n} - \frac{7n}{n}}{\frac{-5n}{n} + \frac{14}{n}} = \lim_{n \to \infty} \frac{2n - 7}{-5 + 0} = -\infty.
\]
Since the term \( 2n \) grows without bound, this limit diverges to \( -\infty \).

### Problem (4)
\[
\lim_{n \to \infty} \frac{2n^2 + 4n}{\sqrt{5n^2 + 14}}
\]
We can factor out \( n \) in the denominator:
\[
\sqrt{5n^2 + 14} = n \sqrt{5 + \frac{14}{n^2}}.
\]
Dividing the numerator and the denominator by \( n \), we have:
\[
\lim_{n \to \infty} \frac{\frac{2n^2}{n} + \frac{4n}{n}}{n \sqrt{5 + \frac{14}{n^2}}} = \frac{2 + 0}{\sqrt{5 + 0}} = \frac{2}{\sqrt{5}}.
\]

### Summary of Results
1. \( \lim_{n \to \infty} \frac{2n^3 - 7n^2 + 8n^5}{5n^5 + 4n^2 - 2} = \frac{8}{5} \)
2. \( \lim_{n \to \infty} \frac{2n^2 - 7}{\sqrt{5n^4 + 14n^2}} = \frac{2}{\sqrt{5}} \)
3. \( \lim_{n \to \infty} \frac{2n^2 - 7n}{-5n + 14} = -\infty \)
4. \( \lim_{n \to \infty} \frac{2n^2 + 4n}{\sqrt{5n^2 + 14}} = \frac{2}{\sqrt{5}} \)

Would you like a detailed explanation of any specific part of these calculations?

Here are some related questions:
1. How can you determine the dominant term in a polynomial limit?
2. What happens to rational functions as \( n \to \infty \) when the numerator's degree is higher than the denominator’s?
3. How do square roots affect the growth rate of terms involving polynomials?
4. What is the impact of a negative coefficient in the denominator on the limit?
5. Why does dividing each term by the highest power of \( n \) simplify limits as \( n \to \infty \)?

**Tip:** When dealing with limits of sequences or functions at infinity, always compare the highest power terms in the numerator and denominator to determine the dominant behavior.

Calculate the limits of sequences provided, justifying or explaining the results:

Learn to calculate the limits of sequences involving rational functions and polynomials, with detailed solutions for each limit. This problem explores key calculus concepts including polynomial growth rates and the behavior of functions at infinity. Suitable for university-level calculus students.

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