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...

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.
