Let $(a_n)$ be a convergent sequence of positive real numbers. Why is the limit nonnegative? Let $(a_n)$ be a convergent sequence of positive real numbers. Why is the limit nonnegative?
My try: For all $\epsilon >0$ there is a $N\in \mathbb{N}$ such that $|a_n-L|<\epsilon$ for all $n\ge N$.  And we know $0< a_n$ for all $n\in \mathbb{N}$, particularly $0<a_n$ for all $n\ge N$. Maybe by contradiction: suppose that $L<0$, then $L<0<a_n$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. Then $0<-L<a_n-L$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. It follows: for all $\epsilon >0$, there is a $N\in \mathbb{N}$ such that $0<|-L|=-L<|a_n-L|<\epsilon$ for all $n\ge N$, which can't be true.
Is my proof ok?
 A: Your proof is a bit confused at the end. But it seems that you would conclude $0<|L|<\epsilon$ for every $\epsilon>0$ and you can a get a contradiction by choosing $\epsilon = |L|/2$.
I propose you nevertheless the following formulation:
Suppose by contradiction that $a_n\geq 0$ for every $n$, $\lim\limits_{n\to\infty} a_n=L$ and $L<0$.
Let $\epsilon = |L|/2>0$, by definition of the limit, there exists $N$ such that $|a_n-L|<\epsilon= |L|/2$ for every $n\geq N$. In particular, this implies that
$$a_N-L<|L|/2=-L/2 \implies a_N<L-L/2=L/2 <0.$$
A contradiction to $a_n\geq 0$ for every $n$.
A: Here is a proof that I believe is more succinct than all of the above. 
Suppose $\lim a_n = a <0$ . 
Let $\epsilon=-a>0$ .
By hypothesis, there exists an $n$ large enough such that $\left|a_n-a\right|<\epsilon =-a$ 
$\Rightarrow a_n-a<-a $
$\Rightarrow a_n <0$ 
Contradiction. 
A: If the limit were negative, say $\ell<0$, there would be at least be a term of the sequence (in fact, infinitely many terms) smaller than $\ell/2$, and thus this term would be negative, which is impossible.
A: Using your notation, if $n\ge N$ then $|a_n-L|<\epsilon$. This absolute-value inequality if equivalent to the inequalities $a_n-\epsilon< L < a_n+\epsilon$. In particular (use only the left-hand inequality and take $n=N$), 
$$
-\epsilon<a_N-\epsilon<L,
$$
Summarizing: $-\epsilon <L$ for each $\epsilon>0$.  This implies that $0\le L$.
A: Let $l < 0$ be the limit of $(a_{n})$. Then there is no $n \geq 1$ such that $|l-a_{n}| < |l|$, a contradiction.
A: See, you  have $$L\lt 0\lt a_{n} \ ,\ \ for\ \ all\ \ n\in N\ $$ Now,take  $\epsilon={{|L|}\over {2}}$. Look  at  the  $\epsilon$-nbd  of  $L$. This  has  no  positive  number at  all in  it, let  alone  the  elements  of  the  sequence  $\{a_{n}\}$.  So  this  contradicts  the  fact  that  $L$  is  the  limit  point  of  the  sequence  $\{a_{n}\}$. 
Hence  $L$  must  be  non-negative.
A: If $a_n>0$ for all $n$, then $\liminf_{n\to\infty} a_n \geqslant 0$, for if not, there would exist an $n$ such that $\inf_{k\geqslant n}a_k<0$. But the $\inf$ of a set of positive numbers cannot be negative, so $\liminf_{n\to\infty} a_n\geqslant0$, and because $a_n$ is convergent, $$\lim_{n\to\infty}a_n=\liminf_{n\to\infty} a_n \geqslant 0.$$
