Big Oh and Little oh notation

Question

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.

Answer 1

1. $2n^3 + 5n^2 -3 = o(n^3)$ 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 $n \to \infty.$

2. $\frac{n+2}{\sqrt{n^2+4}} = O(1)$ 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 $n.$