Prove formally that $\frac {n^2 + 2}{3n^3 - 5n}\to 0$ as $n \to \infty$.

Question

I'm reviewing some Sequences notes from a Mathematics Analysis course I'm taking. I'm finding the beginning of the formal proof below confusing. Some clarity on the following questions would be much appreciated.

Questions

• Why is $n\ge3$ noted? If it is to make $\frac {n^2+2}{3n^3 -5n}$ positive, why not use $n\ge2$?
• How does one arrive at $\frac {n^2+2}{3n^3 -5n} \lt \frac {n^2+2}{2n^3}$? Where does $\frac {n^2+2}{2n^3}$ come from?

Formal Proof

Answer 1

If $n \geqslant 3$ then $n^3 > 5n$ which implies $3n^3 -5n > 2n^3$. This is not true if $n = 2$.

Therefore,

$$\frac{n^2+2}{3n^3 -5n} < \frac{n^2+2}{2n^3}.$$

We also have $2 < 2n^2$ for $n \geqslant 2$ which implies $n^2 + 2 < n^2 +2n^2 = 3n^2$.

Hence if $n \geqslant 3$,

$$\frac{n^2+2}{3n^3 -5n} < \frac{3n^2}{2n^3} = \frac{3}{2n} .$$