(b) If (an) is not eventually positive, and −(an) is not eventually positive, show
that limn→∞ an = 0.

We are tasked with showing that if the sequence \((a_n)\) is neither eventually positive nor eventually negative, then the limit of the sequence as \(n \to \infty\) must be 0.

### Definitions:
- A sequence \((a_n)\) is **eventually positive** if there exists some \(N \in \mathbb{N}\) such that for all \(n \geq N\), \(a_n > 0\).
- A sequence \((a_n)\) is **eventually negative** if there exists some \(N \in \mathbb{N}\) such that for all \(n \geq N\), \(a_n < 0\).

The problem states that \((a_n)\) is **not eventually positive**, and \(- (a_n)\) is **not eventually positive**, meaning:
- There is no \(N_1\) such that \(a_n > 0\) for all \(n \geq N_1\).
- There is no \(N_2\) such that \(a_n < 0\) for all \(n \geq N_2\).

### Proof:

We need to show that under these conditions, \( \lim_{n \to \infty} a_n = 0 \).

1. **Case 1: The sequence oscillates between positive and negative values infinitely often.**

   Since the sequence is neither eventually positive nor eventually negative, for any \(N \in \mathbb{N}\), there will be values of \(a_n\) that are positive and values of \(a_n\) that are negative beyond some point.

   - In other words, there exists a sequence of terms \((a_{n_k})\) that take positive values for some indices \(n_k\), and there exists another sequence \((a_{m_k})\) that take negative values for other indices \(m_k\), with both of these subsequences existing beyond any large index.

2. **Case 2: Convergence of \(a_n\) to 0.**

   If the sequence \((a_n)\) does not eventually become positive or negative, and it keeps alternating between positive and negative values as \(n \to \infty\), the only way for this oscillation to settle is if the terms of the sequence approach zero. 

   - To explain further, assume for contradiction that the sequence converges to some nonzero limit, say \(L \neq 0\). If \(L > 0\), then for sufficiently large \(n\), \(a_n\) should be positive, contradicting the assumption that \((a_n)\) is not eventually positive. Similarly, if \(L < 0\), the terms should eventually be negative, contradicting the assumption that \((a_n)\) is not eventually negative.

   Therefore, the only possibility that fits the condition that \((a_n)\) is neither eventually positive nor negative, is that the sequence oscillates between positive and negative values in such a way that it gets closer and closer to zero as \(n \to \infty\).

Thus, we conclude that \( \lim_{n \to \infty} a_n = 0 \).

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Would you like further details or clarifications on any step?

Here are some related questions to deepen the understanding of this topic:
1. How does the notion of oscillating sequences relate to limits in real analysis?
2. What is the formal definition of a sequence being "eventually positive" or "eventually negative"?
3. Can this result be generalized to other types of sequences, such as complex sequences?
4. How does the behavior of oscillating sequences differ from bounded sequences?
5. What other theorems in real analysis deal with limits of oscillating sequences?

**Tip**: When analyzing limits of sequences that oscillate, look for trends where the sequence values approach a specific number (often zero), even though they do not remain consistently above or below that value.

If (an) is not eventually positive, and −(an) is not eventually positive, show that limn→∞ an = 0.

Learn how to prove that the limit of a sequence (a_n) is 0 as n approaches infinity, given that the sequence is neither eventually positive nor eventually negative. This solution involves the convergence of oscillating sequences in real analysis. Suitable for advanced high school or early college students.

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