<s>[INST] Proof of convergence $\frac{n^2+3n+2}{6n^3+5}$ 
I need to prove that  $a_n = \frac{n^2+3n+2}{6n^3+5}$ converges as $n\to\infty$. I think it will converge to $0$, so I want to prove it.
Let $\epsilon >0$. Then

side-work
$$\left|\frac{n^2+3n+2}{6n^3+5}\right|< \left| \frac{n^2+3n+2n}{6n^3+5}\right| < \left| \frac{n^2+5n}{6n^3}\right|$$ where I used $n \geq 1$ and then I continued $$\left| \frac{n^2+5n}{6n^3}\right| = \left|\frac{n+5}{6n^2}\right|<\left|\frac{6n}{6n^2}\right|$$ using again $n \geq 1$ and finally $$\left|\frac{6n}{6n^2}\right| = \left|\frac{1}{n}\right|=\frac{1}{n} < \epsilon$$

..Then take $n_0 = \max\left\{1, \frac{1}{\epsilon}\right\}$ and $\forall n > n_0$ we have $\left|\frac{n^2+3n+2}{6n^3+5}-0\right|<\epsilon$

Is this proof correct? Because my books give a different one
 [/INST] Unless you have some particular reason for an epsilon-delta proof:
$$\frac{n^2+3n+2}{6n^3+5} = \frac{n^2}{n^3} \cdot \frac{1 + 3/n + 2/n^2}{6+5/n^3} \;\;\to\;\; 0 \cdot \frac{1}{6} \;=\; 0$$
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