Theorem Explanation: Sequences and Convergence I'm currently studying analysis and came across this theorem: 
Let $(x_n)$ be a sequence of real numbers and let $x \in {\Bbb R}$. If $(a_n)$ is a
sequence of positive real numbers with $\displaystyle\lim_{n\rightarrow \infty}{a_n} =0$ and if for some constant $C > 0$ and some $m \in {\Bbb N}$ we have:
$|x_n - x| \leq Ca_n$ $\forall n \geq m$ then:
$\lim{x_n}  = x$ as $ x \to \infty$
I don't understand what the theorem means so I can't even begin to start constructing a proof for it. I've read the statement over and over again, however, I just can't make sense of it. Can anyone please explain what this theorem is saying?
Thank you. 

This is a different version of the squeeze theorem. It is basically saying that since we know that $a_n$ gets really small as $n \to \infty$, if we know that there is a constant $C> 0$ such that $|x-x_n| \leq Ca_n$ (the distance between $x_n$ and $x$ is a multiple of how close $a_n$ is to $0$ since $a_n = |a_n - 0|$) then $x_n$ gets really close to $x$ as $n\to \infty$.
Example: We know that $a_n = \frac{1}{n} \to 0$ as $n\to \infty$. Since 
$$\left|\frac{5n^2 + n}{n^2 + n^{1/2} + 1} - 5\right| \leq \frac{3}{n},$$ (you can check by a calculation) for $n \geq 25 = m$.
If we write $x_n = \frac{5n^2 + n}{n^2 + n^{1/2} + 1}$ and $x = 5$, then
$$|x_n - x| = \left|\frac{5n^2 + n}{n^2 + n^{1/2} + 1} - 5\right| \leq \frac{3}{n} = 3a_n =Ca_n,$$ and so
 the theorem that you are wanting to prove says that $x_n \to x$.
