How do you prove this limit inequality? Let $(a_n)$ and $(b_n)$ be convergent series. Prove that if $a_n \leq b_n$, $(\forall) n \in \mathbb{N}$, then $$ \lim_{n \to \infty} a_n \leq \lim_{n \to \infty}b_n$$
Ok guys, how can you prove this? And if possible, I would like a proof that is not taken out of of the hat. I know the way things works with real analysis and how we usually come up with some interesting proof that require some computation tricks, but please, give me some explanation on how you came up with the idea .
 A: Let $a_n \to a$ and $b_n \to b$. Define $c_n = b_n-a_n$. By hypothesis we know that $c_n\ge0$ for all $n$, then using the limit laws we have $c_n \to c$ where $c= b-a$. It will suffice to show that $c\ge0$. We argue by contradiction, suppose that $c<0$ and let $\varepsilon=|c|/2$. So $|c_n-c|<|c|/2$ for all $n\ge N_\varepsilon$. Thus 
$$c_n<|c|/2+c=c/2<0$$
a contradiction. This contradiction prove that $c=a-b\ge0$ as desired.
Edit: Also you can argue as follows. Given $\varepsilon>0$, there is a $n_0(\varepsilon)$ such that $|b_n-b|< \varepsilon/2$ and $|a_n-a|< \varepsilon/2$ for all $n\ge n_0(\varepsilon)$ at the same time. Thus 
$$a-\varepsilon/2<a_n\le b_n <b+\varepsilon/2$$
which imply $a<b+\varepsilon$  for all $\varepsilon>0$. Thus $a\le b$.
A: Let $\lim_{n\rightarrow \infty}(a_n)=a$ and $\lim_{n\rightarrow \infty}(b_n)=b$. we need to show that $a\leq b$. if not let $a>b$. then $a-b>0$. so take $\epsilon = a-b$ in the definition of convergence of limit. so, $\exists N\in \mathbb{N}$ such that $\forall n\in \mathbb{N}$ and $n\geq N$ we have $|a_n-a|<a-b$ which will give $b<a_n$ $\forall n\in \mathbb{N},\ n\geq N$. Can you arrive at a contradiction now ?
