Simple Limit Question Given: $c\in \mathbb{R} $ and $ {(a_n})$ sequence and l$\in\mathbb{R}$ and
$\forall \,\, \epsilon>0\,\,\exists\ n_0\in\mathbb{N}$ such that $\,\,\forall n>n_0$, $|a_n-l|<c\epsilon$
prove that
$\forall \,\, \epsilon>0\,$,$\,\exists\  n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$
 A: It's important to realize that the epsilons in the two statements:
$\ \ \ \ $1) $\forall \,\, \epsilon>0\,\,\exists\ n_0\in\mathbb{N}$ such that $\,\,\forall n>n_0$, $|a_n-l|<c\epsilon$.
and
$\ \ \ \ $2) $\forall \,\, \epsilon>0\,$,$\,\exists\  n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$.
are not necessarily the same. You could use any symbol you like in either of the statements to play the role of $\epsilon$. It is therefore valid and would perhaps makes things more clear to reformulate the statements as
$\ \ \ \ $1') $\forall \,\, \epsilon'>0\,\,\exists\ n_0\in\mathbb{N}$ such that $\,\,\forall n>n_0$, $|a_n-l|<c\epsilon'$.
and
$\ \ \ \ $2') $\forall \,\, \epsilon>0\,$,$\,\exists\  n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$.
You need to show that 2') holds if 1') holds. 
Hint: 
Note that in order for 1') to hold, we must have $c>0$. Thus,  given any $\epsilon>0$, you can, by choosing $\epsilon'$ appropriately, make $c\epsilon'=\epsilon$.

Solution:
We fix a value of $\epsilon >0$. Then if we set $\epsilon'=\epsilon/c$ we have $\epsilon'> 0$ (recall that $c$ must be positive). We now appeal to  1') to obtain  an $n_0$, so that  $\forall \ n>n_0 $:
$$|a_n-l|<c\epsilon'=c\cdot(\epsilon /c)=\epsilon; $$ and we are done.
