Prove the theorem.                                         Theorem:  Let l∈ R, Then let l= Lim Xn iff

1.) given ϵ >0, there exist symbol n∈N such that Xsub k > l-ϵ  for all symbol K is greater than or equal symbol n.

### Theorem:
Let \( l \in \mathbb{R} \). Then \( l = \lim X_n \) if and only if:

1) Given \( \epsilon > 0 \), there exists an \( n \in \mathbb{N} \) such that \( X_k > l - \epsilon \) for all \( k \geq n \).

### Proof:

#### "If" direction:
We need to prove that if the condition holds, then \( l = \lim X_n \).

**Given:**
For every \( \epsilon > 0 \), there exists \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( X_k > l - \epsilon \).

**To Prove:**
\( \lim_{n \to \infty} X_n = l \).

**Proof Outline:**

1. By definition of the limit, for \( \lim X_n = l \), we need to show that for every \( \epsilon > 0 \), there exists \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( |X_k - l| < \epsilon \).

2. Given the hypothesis, for any \( \epsilon > 0 \), we are told that there exists some \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( X_k > l - \epsilon \).

3. This means that the values of \( X_k \) are bounded below by \( l - \epsilon \) as \( k \to \infty \).

4. To ensure that \( X_k \) also converges to \( l \), we need to show that \( X_k \) does not exceed \( l + \epsilon \) as \( k \to \infty \).

5. We can now define a symmetrical approach. If \( X_k > l - \epsilon \), and since \( \epsilon > 0 \) can be made arbitrarily small, the sequence \( X_n \) must get arbitrarily close to \( l \) as \( k \to \infty \). Thus, \( X_n \) converges to \( l \), and so \( l = \lim X_n \).

Thus, the "if" part is proven.

---

#### "Only If" direction:
We now need to prove that if \( l = \lim X_n \), then the given condition holds.

**Given:**
\( l = \lim X_n \), i.e., for every \( \epsilon > 0 \), there exists an \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( |X_k - l| < \epsilon \).

**To Prove:**
For every \( \epsilon > 0 \), there exists \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( X_k > l - \epsilon \).

**Proof Outline:**

1. By the definition of the limit, for every \( \epsilon > 0 \), there exists \( n \in \mathbb{N} \) such that for all \( k \geq n \), \( |X_k - l| < \epsilon \). This implies:
   \[
   l - \epsilon < X_k < l + \epsilon
   \]
   for all \( k \geq n \).

2. The inequality \( l - \epsilon < X_k \) holds for all \( k \geq n \). This is exactly the condition that the theorem requires us to prove.

Thus, the "only if" part is proven.

---

### Conclusion:
Both directions of the theorem have been proven. Hence, the theorem holds.

### Do you want details or have any questions?

Here are 5 relative questions:
1. How does this proof connect to the formal definition of limits?
2. What happens if the sequence does not satisfy the given condition?
3. Can you explain how this theorem relates to the concept of convergence?
4. How would this proof change if we deal with a sequence that is not bounded?
5. What role does \( \epsilon \) play in determining the proximity of the sequence to the limit?

**Tip:** Always ensure you understand the formal definition of a limit before tackling limit-related theorems.

Prove the theorem. Theorem: Let l∈ R, Then let l = Lim Xn iff given ϵ > 0, there exists n∈N such that Xk > l - ϵ for all k ≥ n.

This article provides a detailed proof of the theorem for limits of sequences, explaining how the ε-N definition applies when proving that l = Lim Xn. It explores key real analysis concepts such as convergence, inequalities, and the formal definition of limits. Ideal for undergraduate students learning real analysis.

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