1. ## Advanced Calculus - Limit Proof

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

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

So far I've gotten:

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

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

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

So far I've gotten:

Given $\epsilon > 0$ then,
$|x_n-x|<\epsilon$
$-\epsilon
$-\epsilon + x
Since $x\in(0,\infty)$ which is an open set there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq(0,\infty)$ and since $x_n\to x$ there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies x_n\in B_{\delta}(x)$. Form you conclusion.

3. Originally Posted by Drexel28
$N\leqslant n\implies x_n\in B_{\delta}(x)$
Could you clarify that part? I'm not sure how you went from left to imply right.

4. 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 $\varepsilon=\frac{x}{2}>0$ so there exists some $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|x_n-x\right|<\frac{x}{2}\implies x+\frac{-x}{2} and so...

5. ## 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 $n \geq K(epsilon)$ from the definition of the $lim(x_n)=x$, let K(epsilon) = M.