Cauchy sequence limit Can someone help me with this question please:

Let $(a_n)_{n\in\mathbb N}$ be a Cauchy sequence, with limit $\ell$. Show that if $N\in\mathbb N_+$, is such that $$n,m\ge N \Rightarrow |a_n-a_m|\le\varepsilon$$ then, for the same $N\in\mathbb N_+$, $$n\ge N \Rightarrow |a_n-\ell|\le\varepsilon.$$


There exists $N$ s.t. $n,m\geq N \Rightarrow |a_n-a_m|\leq \epsilon$. The $m$ can get as large as possible but still the maximum it can get is the limit $l$ which still has to be within epsilon. So that's what we need to show but I am not sure how to do it rigorously. 
 A: Choose any $c > 0$.
Since
$\lim_{n \to \infty} a_n = l$,
for large enough $n$,
$|a_n - l| < c$.
Therefore,
for large enough $n$ and $m$,
$|a_n - l|
= |a_n - a_m + a_m - l|
\le |a_n - a_m| + |a_m - l|
\le \epsilon + c
$.
If $|a_n-l| > \epsilon$
for all large enough $n$,
then there is a $d > 0$
such that
$|a_n-l| \ge \epsilon + d$,
but this is contradicted by choosing,
for example,
$c = d/2$
in
$|a_n - l|
\le \epsilon + c
$.
Therefore $|a_n-l| \le \epsilon$
for large enough $n$.
A: Here is a direct way to see this result. Consider the quantities
$$A^N := \sup_{n,m\geq N}\{|a_n-a_m|\} $$
and
$$B^N := \sup_{n\geq N}\{|a_n-l|\} $$
Notice that the assumption that for all $n,m\geq N$ we have $|a_n-a_m|\leq \epsilon$ implies that $A^N\leq \epsilon$. Now notice that we will be done if we can show that $B^N\leq \epsilon$ since if $k\geq N$ then $|a_k-l|\leq B^N\leq\epsilon$.
The fact that $B^N\leq \epsilon$ will follow immediately by showing that $B^N\leq A^N$. To check this, let $\delta>0$ and choose $M\geq N$ such that if $m\geq M$ then $|a_m-l|\leq \delta$. Then
$$|a_n-l| = |a_n-a_m+a_m-l|\leq |a_n-a_m|+|a_m-l| \leq |a_n-a_m|+\delta$$
Now take the sup of both sides
$$\sup_{n\geq N}|a_n-l|\leq \sup_{n\geq N}|a_n-a_m| +\delta$$
Then since $m\geq M\geq N$, we have
$$\sup_{n\geq N}|a_n-l|\leq \sup_{n\geq N}|a_n-a_m| +\delta\leq \sup_{n,m\geq N}|a_n-a_m| +\delta$$
Now this is true for all $\delta>0$, hence we must conclude that $B^N\leq A^N$.
