Math Problem Statement

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.

Solution

Theorem:

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

  1. Given ϵ>0\epsilon > 0, there exists an nNn \in \mathbb{N} such that Xk>lϵX_k > l - \epsilon for all knk \geq n.

Proof:

"If" direction:

We need to prove that if the condition holds, then l=limXnl = \lim X_n.

Given: For every ϵ>0\epsilon > 0, there exists nNn \in \mathbb{N} such that for all knk \geq n, Xk>lϵX_k > l - \epsilon.

To Prove: limnXn=l\lim_{n \to \infty} X_n = l.

Proof Outline:

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

  2. Given the hypothesis, for any ϵ>0\epsilon > 0, we are told that there exists some nNn \in \mathbb{N} such that for all knk \geq n, Xk>lϵX_k > l - \epsilon.

  3. This means that the values of XkX_k are bounded below by lϵl - \epsilon as kk \to \infty.

  4. To ensure that XkX_k also converges to ll, we need to show that XkX_k does not exceed l+ϵl + \epsilon as kk \to \infty.

  5. We can now define a symmetrical approach. If Xk>lϵX_k > l - \epsilon, and since ϵ>0\epsilon > 0 can be made arbitrarily small, the sequence XnX_n must get arbitrarily close to ll as kk \to \infty. Thus, XnX_n converges to ll, and so l=limXnl = \lim X_n.

Thus, the "if" part is proven.


"Only If" direction:

We now need to prove that if l=limXnl = \lim X_n, then the given condition holds.

Given: l=limXnl = \lim X_n, i.e., for every ϵ>0\epsilon > 0, there exists an nNn \in \mathbb{N} such that for all knk \geq n, Xkl<ϵ|X_k - l| < \epsilon.

To Prove: For every ϵ>0\epsilon > 0, there exists nNn \in \mathbb{N} such that for all knk \geq n, Xk>lϵX_k > l - \epsilon.

Proof Outline:

  1. By the definition of the limit, for every ϵ>0\epsilon > 0, there exists nNn \in \mathbb{N} such that for all knk \geq n, Xkl<ϵ|X_k - l| < \epsilon. This implies: lϵ<Xk<l+ϵl - \epsilon < X_k < l + \epsilon for all knk \geq n.

  2. The inequality lϵ<Xkl - \epsilon < X_k holds for all knk \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 ϵ\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.

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Math Problem Analysis

Mathematical Concepts

Real Analysis
Limits of Sequences
Inequalities

Formulas

|Xn - l| < ϵ
l = Lim Xn

Theorems

Limit Theorem for Sequences

Suitable Grade Level

Undergraduate (First-Year Calculus or Real Analysis)