Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
 A: $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\Leftrightarrow \left|\frac{n^3+3n}{2n^4-n+1}\right|<\epsilon.$$ Now, $4n^3\ge n^3+3n$ and $2n^4-n+1\ge n^4$ for all $n$ positive integer. So $$\left|\frac{n^3+3n}{2n^4-n+1}\right|\le \frac{4n^3}{n^4}=\frac{4}{n},$$ and choosing $n_0=\lfloor 4/\epsilon \rfloor+1$  we have $\dfrac{4}{n}<\epsilon$ if $n\ge n_0,$ as a consequence $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\text{ if }n\ge \left\lfloor \frac{4}{\epsilon} \right\rfloor+1.$$
A: 
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$

Then, since $n^2 \gt 3$ and $2n^3-1 \gt n^3$ for $n \gt 1\,$:
$$
\require{cancel}
\frac{n^3+3n}{2n^4-n} = \frac{\cancel{n}(n^2+3)}{\cancel{n}(2n^3-1)} \lt \frac{n^2+n^2}{n^3} = \frac{2}{n}
$$
You should hopefully be able to take it from there.
A: Let $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\epsilon}{4}$. If $n\geq N$, then we get
$$\begin{align}\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3 \right|&
=\left|\frac{n^3+3n}{2n^4-n+1}\right|\\&=\frac{n^3+3n}{2n^4-n+1}\\
&<\frac{n^3+3n}{2n^4-n}\\
&\leq\frac{n^3+3n}{2n^4-n^4}\\
&\leq\frac{n^3+3n^3}{n^4}\\
&=\frac{4}{n}\leq\frac{4}{N}<\epsilon.
\end{align}$$
Hope it helps.
