To prove the theorem stated in the image, the formal expression states that:

Let $l \in \mathbb{R}$. Then $l = \lim X_n$ if and only if:

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

Proof:

This is essentially a variant of the definition of the limit of a sequence in real analysis.

We need to show that $l = \lim X_n$ if and only if the condition described holds. Here's how we can break it down:

Part 1: $l = \lim X_n \Rightarrow$ the given condition holds.

Hypothesis:

• By the definition of a limit, $l = \lim_{n \to \infty} X_n$ means that for every $\epsilon > 0$, there exists an $n \in \mathbb{N}$ such that: $|X_k - l| < \epsilon \quad \text{for all} \quad k \geq n.$ This implies: $l - \epsilon < X_k < l + \epsilon \quad \text{for all} \quad k \geq n.$ From this, we directly get: $X_k > l - \epsilon \quad \text{for all} \quad k \geq n.$ Thus, the condition is satisfied.

Part 2: The given condition holds $\Rightarrow l = \lim X_n$.

Hypothesis:

• We are given that for every $\epsilon > 0$, there exists an $n \in \mathbb{N}$ such that $X_k > l - \epsilon$ for all $k \geq n$.

We now want to show that this implies $l = \lim X_n$.

We already know that $X_k > l - \epsilon$ for all $k \geq n$, but to fully prove this, we would need to extend this to also show the upper bound $X_k < l + \epsilon$ (i.e., the sequence stays within an $\epsilon$-neighborhood of $l$).

Thus, combining both conditions (lower and upper bounds), we conclude that $X_k$ converges to $l$, and hence $l = \lim X_n$.

Conclusion:

We've shown that both directions hold, proving that $l = \lim X_n$ if and only if the given condition about $X_k$ and $l - \epsilon$ holds.

Would you like more detailed steps or clarification on any part of this proof?

Here are some related questions for further exploration:

1. What is the general definition of a sequence converging to a limit in $\mathbb{R}$?
2. Can we extend this theorem to sequences in other metric spaces?
3. How would the proof change if we considered $X_n$ to be a sequence of complex numbers?
4. What happens if the sequence $X_n$ does not converge to $l$? Can we still apply a similar condition?
5. How does this theorem relate to the Cauchy criterion for convergence?

Tip: When proving limits, always carefully track both lower and upper bounds to ensure the sequence is fully contained within the $\epsilon$-neighborhood of the limit.