Proof or counterexample of a property of limits of sequences Let $a_n$ and $b_n$ be two sequences of positive integers such that
$$\lim_{n \rightarrow \infty} |a_n - b_n| = 0.$$
Must it be the case that 
$$\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 1?$$
I have not been able to come up with a proof or counterexample!  
 A: As other people here stated, it is true for integer numbers.
Indeed we don't need that they be positive, all we need is that $b_n \neq 0$.
For non-integer numbers, set $a_n=2/n$ and $b_n=1/n$.
A: Since $a_n$ and $b_n$ are integers, then the condition $\lim_{n \rightarrow \infty} |a_n - b_n| = 0$ implies that there exists an $N$ for which $a_n = b_n$ for all $n \geq N$. Therefore the claim is true.
A: More generally, if $a_n,b_n$ are positive reals, and $\liminf_{n\to\infty} b_n>0$, then this is true.
This is because $\left|\frac{a_n}{b_n}-1\right| =\frac{1}{b_n}|a_n-b_n|$. 
So letting $\epsilon=\frac{1}{2}\liminf b_n$ (or, if $\liminf b_n=+\infty$, letting $\epsilon=1$) we have that for some $N$,  $b_n>\epsilon>0$ for all $n>N$.
This means that:$$\left|\frac{a_n}{b_n}-1\right|\leq \frac{1}{\epsilon}|a_n-b_n|$$
With $b_n$ positive integers, $\liminf b_n\geq 1$. So you don't even need $a_n$ to be integers.
You can actually generalize even more, allowing for any reals. If $\liminf_{n\to\infty}|b_n|>0$ then $|a_n-b_n|\to 0$ means $\frac{a_n}{b_n}\to 1$.
