Prove that there exists a natural number $K$ such that $a_{n} < b_{n}$ for all $n \geq K$ Given $\displaystyle\lim_{n\rightarrow \infty} a_{n}=a$ and $\displaystyle \lim_{n\rightarrow\infty} b_{n}=b$, and we have $a < b$, how does one prove that there exist a natural number $K$ such that $a_{n} < b_{n}$ for all $n\geq K$?
I can see that intuitively as the sequence $a_{n}$ approaches its limit, it becomes really close to $a$ and logically that implies that eventually the value of $a_{n}$ will be smaller than $b_{n}$ when $n$ is sufficiently large, however I can't seem to find a mathematically sound argument to prove my point. I'm thinking if it's possible to solve this question using the epsilon-K thingy, but I kinda have no idea how to start on my proof.
Could someone point me in the right direction?
 A: Hint: Let $\epsilon = b-a$ be the distance between $a$ and $b$. By definition of the limit, for all sufficiently large $n$, the value $a_n$ lies strictly within a distance of $\epsilon/2$ from $a$. Similarly, for all sufficiently large $n$, the value $b_n$ lies strictly within a distance of $\epsilon/2$ from $b$.
Drawing a simple diagram should help you understand it better.
ADDED LATER: First note that $a+\frac{\epsilon}{2} = \frac{a+b}{2} = b-\frac{\epsilon}{2}$. By assumption, there exist $K_1,K_2 >0$ such that
$$\textstyle n \geq K_1 \ \Longrightarrow \ |a_n-a| < \frac{\epsilon}{2} \quad \text{and} \quad n \geq K_2 \ \Longrightarrow \ |b_n-b| < \frac{\epsilon}{2}.$$
Let $K = \max\{K_1,K_2\}$. Then
\begin{align*}
n \geq K \quad & \Longrightarrow \quad \textstyle |a_n-a|< \frac{\epsilon}{2} \quad \text{and} \quad |b_n-b|< \frac{\epsilon}{2} \\
& \Longrightarrow \quad \textstyle -\frac{\epsilon}{2} < a_n-a < \frac{\epsilon}{2} \quad \text{and} \quad -\frac{\epsilon}{2} < b_n-b < \frac{\epsilon}{2} \\
& \Longrightarrow \quad \textstyle a-\frac{\epsilon}{2} < a_n < a+\frac{\epsilon}{2} \quad \text{and} \quad b-\frac{\epsilon}{2} < b_n < b+\frac{\epsilon}{2} \\
& \Longrightarrow \quad\textstyle a_n<a+\frac{\epsilon}{2} = \frac{a+b}{2} = b-\frac{\epsilon}{2} < b_n.
\end{align*} 
A: Lemma: Let $u_n$ be a sequence of real numbers converging to a positive real number, then there exists an $N$ such that for all $n>N$, $u_n>0$.
Proof:
Let $u_n\rightarrow u$, choose $\varepsilon=\frac{u}{2}>0$, then there exists an integer $N$ such that for $n>N$, $|u_n-u|<\frac{u}{2}\Rightarrow 0<\frac{u}{2}<u_n<\frac{3u}{2} \qquad \square$

Use the lemma for $u_n=b_n-a_n$.
