{Proof Check} Let $x_n$ be a convergent sequence. if $x_n \geq 0$ $\forall n \in \mathbb{N}$ then $\lim x_n \geq 0$ Please check and point out if there are logical flaws in my argument. I'm learning analysis but I'm not that confident yet.
Proof: Suppose $x_n \geq 0 $ $\forall n \in \mathbb{N}$ and let $\lim x_n = x &lt; 0$. Then $|x| &gt; 0$. Take $\varepsilon = |x|$ then $\exists N \in \mathbb{N}$ s.t. $\forall n \geq N$,
$$|x_n - x| \leq |x_n| + |x| &lt; \varepsilon = |x|$$
$$0 \leq |x_n| = x_n &lt; |x| - |x| = 0$$
This is a contradiction, thus $x \geq 0$.
You have the right value of $\varepsilon$ (and any value smaller than that but still positive can also serve).
But how did you conclude that $|x_n| + |x|&lt;\varepsilon \text{?}$
I'd write $|x_n-x| = x_n-x$ since $x_n\ge0$ and $-x\ge0,$ and then $x_n-x\ge -x = \varepsilon.$ Thus no matter how big $n$ gets, $|x_n-x|$ never gets less than $\varepsilon.$
The proof seems correct, you showed there is an $\varepsilon$ for which no $N$ exists. The only thing that seems a bit unorthodox is using $x$ as the variable for the limit instead of the more common $L$.