If $a_n$ and $b_n$ are converging sequences, and for all $n\in\mathbb Z+$, $a_n \leq b_n$, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$. I am having a bit of trouble answering this question. If $a_n$  and $b_n$  are converging sequences, and for all $n \in\mathbb Z+$, $a_n$ ≤ $b_n$, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.
Letting $\epsilon>0$, 
I began with showing that there  exists $N_1\in \mathbb N$ such that for all $n \in \mathbb N$, if $n>N_1$, then $|a_n - L|< \epsilon$.
Similarly exists $N_2\in \mathbb N$ such that for all $n \in \mathbb N$, if $n>N_2$, then $|b_n - M|< \epsilon$.  
However from here I do not know how to proceed. Do I use the triangle inequality somehow?

By contradiction,  if $L\gt M$, let $\epsilon=\frac{L-M}2\gt0$.  Then for $n\gt \operatorname{max}(N_1,N_2)$, we have $a_n\gt L-\epsilon=\frac{L+M}2= M+\epsilon\gt b_n$.
