Since this (probably homework) questions is some days old, I think I can answer this question in short.
The wikipedia page (and hopefully your textbook and/or notes too) says
f(n) ∈ O(g(n)) ⇔ lim sup |f(n)/g(n)| < ∞
f(n) ∈ Ω(g(n)) ⇔ lim sup |f(n)/g(n)| > 0
f(n) ∈ Θ(g(n)) ⇔ f(n) ∈ O(g(n)) and f(n) ∈ Ω(g(n))
To prove the left side you can prove the right side.
n² ∈ O(n³)
is true, due to
lim sup |n²/n³| = lim (n²/n³) = lim (1/n) = 0 < ∞
n² ∈ Ω(n³)
is false, due to
lim sup |n²/n³| = lim (n²/n³) = lim (1/n) = 0
2ⁿ ∈ Θ(2n+1)
is true, due to
0 < lim sup |2ⁿ/2<sup>n+1</sup>| = lim (2ⁿ/(2⋅2ⁿ) = lim (1/2) = 1/2 < ∞
n! ∈ Θ((n+1)!)
is false, due to
lim sup |n!/(n+1)!| = lim (n!/((n+1)⋅n!) = lim (1/(n+1)) = 0
Notice: All limits holds for n → ∞
.