Using the epsilon- N definition of the limit verify that: $ \lim_{n\to\infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3}$
Here is what I have so far:
$\mathrm{x}_\mathrm{n}=\frac{n}{\sqrt{9n^2+4n+3}},$ $a=\frac{1}{3}$
$ \forall{\epsilon}$ $\exists{\mathrm{N}_\mathrm{\epsilon}}$ $|$ $\forall{n}\in{\mathbb{N}}$ $n>\mathrm{N}_\mathrm{\epsilon} \Rightarrow|\mathrm{x}_\mathrm{n}-a|<\epsilon$
$|\frac{n}{\sqrt{9n^2+4n+3}}-\frac{1}{3}|\Rightarrow$
Then I applied the got a common denominator and applied the algebraic formula $a-b=\frac{a^2-b^2}{a+b}$ and ended up with the following but I cannot see how I should simplify it any further before I analyze to verify the inequality is less than epsilon. Here is where I am stuck in the simplification:
$$|\frac{4n+3}{9n\sqrt{9n^2+4n+3}+3(9n^2+4n+3)}|$$

I can only guess that the task is about showing that $$ \lim_{n \to \infty} \frac{n}{\sqrt{9n^2+4n+3}} = \frac{1}{3} $$
Here's one way:
It holds that $\frac{n}{\sqrt{9n^2+4n+3}} = \frac{n}{\sqrt{n^2\left(9+\frac{4}{n}+\frac{3}{n^2}\right)}} = \frac{n}{\sqrt{n^2}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}} $
Since $n \ge 1$ it holds that 
$ \frac{n}{\sqrt{n^2}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}}  =\frac{1}{\sqrt{9+\frac{4}{n}+\frac{3}{n^2}}}$
which converges $\to \frac{1}{3}$ for $n \to \infty$
Edit:
This is another approach using $\varepsilon$ kriteria:
We want to show that for every $\varepsilon > 0 $ there is an $n_0 \in \mathbb{N}$ such that $$ \vert \frac{n}{\sqrt{9n^2+4n+3}} -\frac{1}{3} \vert < \varepsilon $$ for all $n \ge n_0$.
It holds that $ \vert \frac{1}{2n} \vert < \frac{\varepsilon}{2} $
now begin with
$$ \left\vert \frac{n}{\sqrt{9n^2+4n+3}} -\frac{1}{3} \right\vert = \left\vert \frac{3n -\sqrt{9n^2+4n+3}}{3\sqrt{9n^2+4n+3}} \right\vert = \left\vert \frac{9n^2-(9n^2+4n+3)}{3\sqrt{9n^2+4n+3}(3n+\sqrt{9n^2+4n+3})} \right\vert = \left\vert \frac{-4n-3}{9n\sqrt{9n^2+4n+3}+3(9n^2+4n+3)} \right\vert < \left\vert \frac{-4n-3}{9n^2} \right\vert = \vert -1 \vert \left\vert\frac{4n+3}{9n^2} \right\vert = \left\vert\frac{4n+3}{9n^2} \right\vert $$
Now making use of the triangle inequality leads to 
$$\left\vert\frac{4n+3}{9n^2} \right\vert \le \left\vert\frac{4n}{9n^2} \right\vert + \left\vert\frac{3}{9n^2} \right\vert \le \left\vert\frac{4}{9n} \right\vert + \left\vert\frac{3}{9n} \right\vert < \left\vert\frac{1}{2n} \right\vert + \left\vert\frac{1}{2n} \right\vert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon  $$
I'm sure there are better or at least more elegant solutions but i think it suffices the task.
