Suppose $\lim_{n\to\infty} a_n = a$ and $\lim _{n\to\infty} b_n = b$. Prove $a \leq b$. 
Question:
Let $(a_n)$ and $(b_n)$ be sequences with $a_n ≤ b_n,\ ∀n$. Suppose $\lim_{n\to\infty} a_n = a$ and $\lim_{n\to\infty} b_n = b$. Prove $a \leq b$.
[Hint: Assume for contradiction, that $a > b$ and use the deﬁnition of convergence with $\varepsilon = \frac{a-b}2$ to produce an $n$ with $a_n > b_n$]

My Work:
By contradiction 
Let $(a_n)$ and $(b_n)$ be sequences with $a_n  > b_n$
Suppose $\lim _{n\to\infty} a_n = a$ and $\lim_{n\to\infty}b_n = b$.
let $\varepsilon=\frac{a-b}2 >0$ , there is a real number $N$ such that if $n > N$, then \begin{align}|a_n+b_n-a-b|&=|a_n-a+b_n-b|\\
&<|a_n-a|+|b_n-b|\\
&=\frac a2+\frac b2\\
&=\varepsilon\end{align}
I am not sure how $\varepsilon=\frac{a-b}2$ and how can I show $a > b$?
Please, help me with that thank you.
 A: You wrote:
$$
\underbrace{|a_n+b_n-a-b|=|a_n-a+b_n-b|<|a_n-a|+|b_n-b|=(a/2)+(b/2)=ε}_\text{Some parts of this are wrong.}
$$
\begin{align}
& |a_n+b_n-a-b| \\[10pt]
= {} &|a_n-a+b_n-b| \\[10pt]
< {} & |a_n-a|+|b_n-b| & & \longleftarrow \text{This should say “}\le\text{''}. \\[10pt]
= {} & \frac a 2 + \frac b 2 & & \longleftarrow \text{This part is completely wrong.} \\[10pt]
= {} & ε & & \longleftarrow \text{So is this.}
\end{align}
Where you have $\text{“}= \dfrac a 2 + \dfrac b 2\text{ ''}$ you need $\text{“} < \varepsilon+ \varepsilon\text{''}.$
Then on the next line you can have $\text{“} = a-b\text{''}.$ And go on from there.
A: Let $t_n=b_n-a_n$ then for all $n$ we have $t_n\geq0$ and 
$$\lim_{n\to\infty} t_n = \lim_{n\to\infty} b_n -a_n= b-a\geq0$$
otherwise if $b-a<0$, for $\varepsilon=\dfrac{a-b}{2}>0$, there exist $N_0$ such that $|t_n-(b-a)|<\dfrac{a-b}{2}$ for all $n\geq N_0$,
but this shows $t_n<\dfrac{a-b}{2}+(b-a)=\dfrac{b-a}{2}<0$ contradiction with 
$t_n\geq0$, for all $n$.
