# Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly from the definition of limit.

One of my homework questions is:

Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly from the definition of limit.

In trying to follow: Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1} = 0$ from the definition , i have so far:

Proof:

It must be shown that for any $\epsilon>0$, there exists an integer $N$ such that $\left|\frac{2n^2+2}{3n^3+1}-0\right| <\epsilon$ whenever $n>N$.

$\left|\frac{2n^2+2}{3n^3+1}-0\right|=\frac{2n^2+2}{3n^3+1}>\frac{2n^2}{3n^3}=\frac{2}{3n}$.

Because in the problem im given, the left side of the inequality that im trying to simplify i guess, is instead greater than the right side, im stuck.

And i dont understand the how or why behind the algebraic manipulations that the original author and one of the answers performs, in the linked question:

$$\left|\frac{n}{n^2 + 1}\right| < \epsilon \text{ whenever }x \gt M.$$

$$n \lt \epsilon(n^2 + 1)$$

$$n \lt \epsilon n^2 + \epsilon$$

Truly do feel clueless at this point!

• You want to get an upper bound, not a lower bound. – marty cohen Apr 16 '15 at 23:29

Let $\epsilon>0$ be given and let $N>\frac{4}{3\epsilon}$. If $n>N$, then $n>\frac{4}{3\epsilon}$ or $$\epsilon>\left|\frac{4}{3n}\right|$$
$$\epsilon>\left|\frac{4n^2}{3n^3}\right|$$ $$\epsilon>\left|\frac{2n^2+2n^2}{3n^3}\right|\geq\left|\frac{2n^2+2}{3n^3}\right|>\left|\frac{2n^2+2}{3n^3+1}\right|=\left|\frac{2n^2+2}{3n^3+1}-0\right|$$
• Why $N>\frac{4}{3\epsilon}$, more specifically why $\frac{4}{3}$? – Churning Butter Apr 17 '15 at 0:10