Show that $\lim_{n \longrightarrow \infty} \frac{3n^2+2n}{4+3n^2}=1$ I need show that $\forall \, \epsilon > 0$, exist $M \in \mathbb N$ such that, $\left|\frac{3n^2+2n}{4+3n^2}-1\right| < \epsilon$ if $n \geq M$.
Let's consider
$\left|\frac{3n^2+2n}{4+3n^2}-1\right|=\left|\frac{2n-4}{4+3n^2}\right|<\left|\frac{2n-4}{3n^2}\right|=\left|\frac{2}{3n}-\frac{4}{3n^2}\right|\leq \left|\frac{2}{3n^2}-\frac{4}{3n^2}\right|=\frac{2}{3n^2}<\epsilon \Leftrightarrow n>\sqrt{\frac{2}{3\epsilon}}$
Then, I must take $M = \left\lceil \sqrt{\frac{2}{3\epsilon}}\right\rceil+1 \in \mathbb{Z}$?
Thanks.
 A: $$\lim_{n \to \infty} \frac{3n^2 + 2n}{3n^2 + 2} = \lim_{n \to \infty} \frac{3 + 2/n}{3+2/n^2} = \frac{3}{3} = 1$$
$\epsilon$-proof:
$$ \left|\frac{3n^2 + 2n}{3n^2 + 2} - 1\right| = |\frac{2n - 2}{3n^2 + 2}| < \epsilon \Rightarrow |2n-2| < 3\epsilon n^2 + 2\epsilon$$
The absolute value can be removed since both sides are nonnegative when $n > 1$.
$$3\epsilon n^2 -2n + 2\epsilon + 2 > 0$$
This is a parabola with the opening facing up. Using the quadratic formula, 
$$n > \frac{2 + \sqrt{2^2 - 4 \cdot 3\epsilon \cdot (2\epsilon + 2)}}{6\epsilon} = \frac{1 + \sqrt{1 - 6\epsilon - 6\epsilon^2}}{3\epsilon} = M$$
$\pm$ is changed to $+$ since the other root is negative.
A: If $n \geq 1$, then
$$
\bigg| \frac{3n^{2}+2n}{4+3n^{2}} - 1 \bigg| 
=
\bigg| \frac{2n-4}{4+3n^{2}} \bigg|
\leq
\frac{2n}{4+3n^{2}} + \frac{4}{4+3n^{2}}
<
\frac{2}{n} + \frac{4}{n^{2}};
$$
taking any $\varepsilon > 0$, if $n \geq \max \{ \lceil \frac{4}{\varepsilon} \rceil + 1, \lceil \sqrt{\frac{8}{\varepsilon}} \rceil + 1 \},$
then 
$$
\frac{2}{n} + \frac{2}{n^{2}} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
$$
