Proving $\{x_n\}$ converges to $a$ when $|x_n-a|\le Cb_n$ for large $n$ and $C$ is a positive constant. If $\{b_n\}$ is a sequence of nonnegative numbers that converges to $0$, and $\{x_n\}$ is a real sequence that satisfies $|x_n-a|\le Cb_n$ for large $n$, where $C$ is a fixed positive constant, prove that $x_n$ converges to $a$.
I understand this is a simple question. I feel that I am missing a very simple detail because my attempt isn't consistent with the hint that is given (or, I just don't understand what I'm doing). I realize problems have many different ways they can be solved. I am new to sequences and I am unsure if what I did makes sense or not.
The hint provided is that any $\epsilon$, including $\large\frac{\epsilon}{c}$, when, $c\gt 0$, works for the definition: A sequence of real numbers $\{x_n\}$ is said to converge to a real number $a\in \mathbb R$ if and only if for every $\epsilon\gt 0$ there is an $N\in \mathbb N$ such that $n\ge N \to |x_n-a|\lt \epsilon$.
My Attempt:
If $b_n$ converges to $0$, then $|b_n|\to 0$ as $n\to \infty$
$|b_n|\to 0$
$C \cdot|b_n|\to 0\cdot C$
$C|b_n|\to 0$
$|x_n-a|\le Cb_n$
Since, $Cb_n\to 0$, then $|x_n-a|\to 0$.
By definition, if $|x_n-a|\to 0$ as $n\to \infty$, then $x_n$ converges to $a$.
Any input would be great.

The idea is good but you need to be more rigorous and stick to the definition:
Let $\epsilon > 0$, then 
$$\lim_{n \to \infty}b_n = 0$$
implies that there is some $N \in \mathbb{N}$ such that $$b_n=|b_n|<\frac{\epsilon}{ C}$$ for every $n >N$. It follows that $$|x_n-a|\leq Cb_n < C\frac{\epsilon}{ C} = \epsilon$$ for every $n>N$, i.e. $$\lim_{n \to \infty}x_n = a.$$
