# Sequences close to each other have the same limit.

Let ${a_n}$ and ${b_n}$ be two sequences with following property: $\forall \epsilon >0,\ \exists n_0\in \mathbb{N}$ such that $n>n_0 \implies |a_n-b_n| <\epsilon$.

How do I show that $$\lim_{n \to \infty}a_n=L \iff \lim_{n\to \infty}b_n=L?$$

I tried to solve this problem by cauchy sequence but I unable to prove that this is my attempt

$|a_m-a_n|<e/2----(1)$

$|b_m-b_n |<e/2----(2)$

but I can't go further.

hint: $|b_n-L| \leq |b_n-a_n|+|a_n-L|$, and $|a_n-L| \leq |a_n-b_n|+|b_n-L|$.