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Chapter 3: Problem 10

Prove that if \(\lim \left(x_{n}\right)=x\) and if \(x>0\), then there exists a natural number \(M\) such that \(x_{n}>0\) for all \(n \geq M\).

Short Answer

Expert verified
By the epsilon-delta definition of a limit, given \(x > 0\), we can set epsilon to \(\frac{x}{2}\) which is a positive number. Consequently, there exists a natural number \(N\) such that for all \(n \geq N\), \(|x_n - x| < \frac{x}{2}\). This implies that for all determined \(n \geq N\), \(x_n\) is positive, completing the proof.

Step by step solution

01

Understand the Epsilon-Delta Definition of a Limit

This step is to recall the epsilon-delta definition of a limit. According to this definition, \(\lim \left(x_{n}\right)=x\) means that for any epsilon > 0 ,there exists a natural number \(N\) such that for all \(n \geq N\), we have \(|x_n - x| < \epsilon\). This is the definition we'll use to prove the statement.
02

Set Up the Inequation

Now, given the condition that \(x > 0\), we can set \(\epsilon = \frac{x}{2}\) which is positive. By the definition of the limit, there exists a natural number \(N\) such that if \(n \geq N\), then \(|x_n - x| < \frac{x}{2}\).
03

Simplify the Inequation

This inequality means \(x - \frac{x}{2}< x_n < x + \frac{x}{2}\), or equivalently, \(\frac{x}{2} < x_n\). Therefore, we set \(M = N\) such that \(x_{n}>0\) for all \(n \geq M\).

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