A pair of mutually "bounding" sequences. Given $({a_n})_{n=1}^{\infty}$, $({b_n})_{n=1}^{\infty}$ convergent sequences
and where
$$\{n\in\mathbb{N}\mid a_n\le b_n\}\quad\text{and}\quad\{n\in\mathbb{N}\mid b_n\le a_n\}$$
are both unbounded, prove that 
$$\lim \limits_{n\to \infty}a_n=\lim \limits_{n\to \infty}b_n$$
I would like to know how I can prove it using simple calculus theorem(I only know the definition of limit, arithmetics of limits and the Squeeze Theorem). 
Thank you very much for your time and help.
 A: If you want to proceed from basics, meaning the $\epsilon$-$N$ definition of limit,  let the limits of our sequences be $a$ and $b$ respectively. 
We show that we cannot have $a<b$, and that we cannot have $a>b$. To show that $a<b$ is not possible, let $\epsilon=(b-a)/3$.  There is an $N$ such that if $n >N$ then $a_n$ is within $\epsilon$ of $a$, and $b_n$ is within $\epsilon$ of $b$. But then we cannot have $a_n \ge b_n$, contradicting the fact that the set of $n$ such that $a_n \ge b_n$ is unbounded.
A: Here's a somewhat different take that uses the arithmetic of sequences in a central way. 
Since $\lim_{n\to\infty} a_n =a$ and  $\lim_{n\to\infty} b_n = b$, we have by arithmetic of sequences that $\lim_{n\to\infty} \{a_n - b_n\} = a-b$. By hypothesis, $\{a_n-b_n\}$ has infinitely many positive terms and infinitely many negative terms. If $a-b$ is positive, then the negative terms can't get arbitrarily close to $a-b$ as they would have to do. 
Formally, for every $N$ there is some $n\ge N$ for which $a_n-b_n<0$ and thus $|(a-b) - (a_n-b_n)| \ge a-b$. Thus $ \{a_n-b_n\}\not\to a-b$, contradiction.
On the other hand, if $a-b$ is negative, then the positive terms can't get arbitrarily close to $a-b$, again a contradiction. (The formal statement is almost exactly like the one above.) Therefore $a-b=0$.
