# Advanced Calculus - Limit Proof

• Feb 3rd 2010, 04:35 PM
Luxury
Prove that if $\displaystyle lim(x_n)=x$ and if $\displaystyle x>0$, then there exists a natural number $\displaystyle M$ such that $\displaystyle x_n>0$ for all $\displaystyle n \geq M$.

My thoughts
Intuitively, this makes sense. If as $\displaystyle n -> \infty, x_n=x>0$ then this means that at some point $\displaystyle n \geq M$ where $\displaystyle x_n>0$.

So far I've gotten:

Given $\displaystyle \epsilon > 0$ then,
$\displaystyle |x_n-x|<\epsilon$
$\displaystyle -\epsilon<x_n-x<\epsilon$
$\displaystyle -\epsilon + x <x_n<\epsilon + x$
• Feb 3rd 2010, 04:44 PM
Drexel28
Quote:

Originally Posted by Luxury
Prove that if $\displaystyle lim(x_n)=x$ and if $\displaystyle x>0$, then there exists a natural number $\displaystyle M$ such that $\displaystyle x_n>0$ for all $\displaystyle n \geq M$.

My thoughts
Intuitively, this makes sense. If as $\displaystyle n -> \infty, x_n=x>0$ then this means that at some point $\displaystyle n \geq M$ where $\displaystyle x_n>0$.

So far I've gotten:

Given $\displaystyle \epsilon > 0$ then,
$\displaystyle |x_n-x|<\epsilon$
$\displaystyle -\epsilon<x_n-x<\epsilon$
$\displaystyle -\epsilon + x <x_n<\epsilon + x$

Since $\displaystyle x\in(0,\infty)$ which is an open set there exists some $\displaystyle \delta>0$ such that $\displaystyle B_{\delta}(x)\subseteq(0,\infty)$ and since $\displaystyle x_n\to x$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies x_n\in B_{\delta}(x)$. Form you conclusion.
• Feb 3rd 2010, 05:10 PM
Luxury
Quote:

Originally Posted by Drexel28
$\displaystyle N\leqslant n\implies x_n\in B_{\delta}(x)$

Could you clarify that part? I'm not sure how you went from left to imply right.
• Feb 3rd 2010, 05:27 PM
Drexel28
Quote:

Originally Posted by Luxury
Could you clarify that part? I'm not sure how you went from left to imply right.

You may not be familiar with the open ball notation. Well, how about letting $\displaystyle \varepsilon=\frac{x}{2}>0$ so there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies \left|x_n-x\right|<\frac{x}{2}\implies x+\frac{-x}{2}<x_n$ and so...
• Feb 21st 2010, 11:30 PM
Jeanne
Lim(x_n)=x and x>0
Since x>0, it is a little easier to simply let x=epsilon.
Substitute in x for epsilon in you inequality.
And since $\displaystyle n \geq K(epsilon)$ from the definition of the $\displaystyle lim(x_n)=x$, let K(epsilon) = M.