Prove a Limit Using the $\varepsilon - N$ definition I'm having trouble proving the following limit using the $\varepsilon-N$ definition:
$$\lim_{n \to \infty} \frac{3n}{\sqrt{9n^2 + 11n + 4}} = 1.$$
Here is my work so far.
Take arbitrary $\varepsilon$. We need to prove the existence of a natural number, $N$, such that for all $n \ge N$, $$ | x_n - 1| < \varepsilon.$$
$x_n = \frac{3n}{\sqrt{9n^2 + 11n + 4}}$ and since $x_n < \frac{3n}{\sqrt{9n^2}} = 1$, the equation becomes: $$1 - \frac{3n}{\sqrt{9n^2 + 11n + 4}} < \varepsilon.$$
Rewriting gives: $$\frac{3n}{\sqrt{9n^2 + 11n + 4}} > 1 - \varepsilon.$$
Clearly, when $\varepsilon \ge 1$, the statement is true since the RHS of the inequality becomes non-positive and obviously, the RHS must be positive.
However, I'm having trouble proving the statement when $0 < \varepsilon < 1$.
My idea so far has been to take the square:
$$\frac{9n^2}{9n^2 + 11n + 4} > 1 - 2\varepsilon + \varepsilon^2.$$Then, by taking the reciprocal of both sides, we obtain:
$$\frac{9n^2 + 11n + 4}{9n^2} < \frac{1}{1 - 2\varepsilon + \varepsilon^2}.$$
From here, I thought it might be smart to use the idea that for any real number, $x$, there exists a natural number, $m$, such that $\frac{1}{m} < x$.
Specifially, this would give us $m$ such that: $$\frac{1}{m} < \frac{1}{1 - 2\varepsilon + \varepsilon^2}$$ in which we case we could try to solve for $n$ in terms of $m$.
However, this solution is very complicated and long and bashy. There has to be a more elegant way to prove this limit right?
A pattern I've seen in a lot of limit problems involving square roots is the usage of $$a -b = \frac{a^2 - b^2}{a+b}$$ but I can't see how to apply that here either.
Any ideas or tips for a better solution would be greatly appreciated!

Let us assume that $n\geq n_{\varepsilon}$. Then it results that
\begin{align*}
\left|\frac{3n}{\sqrt{9n^{2} + 11n + 4}} - 1\right| & = \left|\frac{3n - \sqrt{9n^{2} + 11n +  4}}{\sqrt{9n^{2} + 11n + 4}}\right|\\\\
& \leq\left|\frac{3n - \sqrt{9n^{2} + 11n + 4}}{3n}\right|\\\\
& = \left|\frac{11n + 4}{3n[3n + \sqrt{9n^{2} + 11n + 4}]}\right|\\\\
& \leq\left|\frac{11n + 4}{3n(3n + 3n)}\right|\\\\
& = \frac{11n + 4}{18n^{2}}\\\\
& = \frac{11}{18n} + \frac{2}{9n^{2}}\\\\
& \leq \frac{11}{18n_{\varepsilon}} + \frac{2}{9n^{2}_{\varepsilon}} := \varepsilon
\end{align*}
Then you can take the ceiling function of the corresponding solution.
