Big Oh and Little oh notation I have to prove or disprove the following:
$$2n^3 + 5n^2 -3 = o(n^3)$$
$$\frac{n+2}{\sqrt{n^2+4}} = O(1)$$
My attempt at the first question
$$2n^3 + 5n^2 - 3 \leq c |g(x)|$$
$$2n^3+5n^2-3 \leq c |2n^3+5n^3-3n^3|$$
$$2n^3+5n^2-3 \leq c |4n^3|$$
$$2n^3+5n^2-3 \leq 4 |n^3|$$
evaluating when n = 1
$$4 \leq 4$$
when n = 2
$$53 \leq 32$$
hence $$fn \neq o(n^3)$$
My attempt at the second question
$$f_n = \frac{n+2}{\sqrt{n^2+4}} $$
$$ \lim_{n \to -\infty} f_n = \lim_{n \to -\infty} \frac{n+2}{\sqrt{n^2+4}} $$
$$ \lim_{n \to -\infty} f_n = \lim_{n \to -\infty} \frac{n}{\sqrt{n^2}} $$
$$ \lim_{n \to -\infty} f_n =  1 $$
Therefore $$f_n = O(1)$$
can anyone just verify my solution and guide me as to whatever I did wrong.

*

*$2n^3 + 5n^2 -3 = o(n^3)$ means $ \frac{2n^3+5n^2-3}{n^3} \to 0$ as $n \to \infty.$
But this is fals, since $ \frac{2n^3+5n^2-3}{n^3} \to 2$ as $n \to \infty.$


*$\frac{n+2}{\sqrt{n^2+4}} = O(1)$ means that the sequence $(\frac{n+2}{\sqrt{n^2+4}})$ is bounded. 


This is true, since $0 \le \frac{n+2}{\sqrt{n^2+4}} \le 4$ for all $n.$
