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...


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.
