im taking my first course in real analysis this summer and I would like some feedback on proof writing. Thank you.
Let $\{x_n\}_n$ be a sequence of real number such that $x_n>0$ for all $n \in \textbf{}N$ and $\lim_{x\to\infty}x_n=l>0.$
1) Let $r<l$. Show that there exist an $N(\epsilon) \in \textbf{N}$ such that: $$n \geq N(\epsilon) \implies x_n > r$$ .
2) Show that $\inf \{ x_n \ | \ n \in \textbf{N} \} > 0$
$\textbf{Question 1.}$
We have: $$\lim_{x\to\infty}x_n=l>0.$$
The definition of the limit say that : $$\forall \epsilon>0, \ \ \ \exists N(\epsilon) \in \textbf{N} : |x_n - l|<\epsilon \ , \ \ \ \forall n>N(\epsilon).$$
Let $\epsilon = l - r$ and then : $$|x_n - l |< \epsilon.$$ $$\iff -\epsilon < x_n - l< \epsilon \ \ , \ \ \ \ \forall n>N(\epsilon). $$ $$\iff l-\epsilon<x_n<l+\epsilon \ \ , \ \ \ \ \ \forall n>N(\epsilon). $$ $$\iff l-(l - r)<x_n<l+\epsilon \ \ , \ \ \ \ \ \forall n>N(\epsilon). $$ $$\iff r<x_n<l+\epsilon \ \ , \ \ \ \ \ \ \ \ \ \ \ \ \forall n>N(\epsilon). $$
$\textbf{Question 2.}$ I'm not sure if I understand the question correctly.
Proof by contradiction:
Suppose that $$\inf\{x_n \ | \ n \in \textbf{N} \} \leq 0$$ and $${x_n}>0 \ \forall n \in \textbf{N}$$
Clearly we have an contradiction.
We conclude that $$\inf\{x_n \ | \ n \in \textbf{N} \} \geq 0.$$
Any feedback is appreciated. Thank you very much.