> 
> **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.