Analysis I MT22, Limit inequality theorem

What contradiction do you use to prove that limits preserve weak inequalities, i.e. if $a _ n \to L$, $b _ n \to M$ and $a _ n \le b _ n$ for all $n$ then $L \le M$?

When proving that limits preserve weak inequalities, i.e. if $a _ n \to L$, $b _ n \to M$ and $a _ n \le b _ n$ for all $n$ then $L \le M$, what value of $\epsilon$ do you take in $ \vert a _ n - L \vert < \epsilon$ and $ \vert b _ n - L \vert < \epsilon$?

When proving that limits preserve weak inequalities, i.e. if $a _ n \to L$, $b _ n \to M$ and $a _ n \le b _ n$ for all $n$ then $L \le M$, you use $\epsilon = \frac{1}{2}(L - M)$. What expression are you trying to show true for a contradiction?

When proving that limits preserve weak inequalities, i.e. if $a _ n \to L$, $b _ n \to M$ and $a _ n \le b _ n$ for all $n$ then $L \le M$, you “expand out” $ \vert a _ n - L \vert < \epsilon$ and $ \vert b _ n - M \vert < \epsilon$ into what?