Prove or disprove the following statement using the definition of big-Θ: NOTE: I am not provign big O here I am proving big-Θ
Prove or disprove the following statement using the definition of big-Θ:
$$n^2−4n = Θ(2^n)$$
so, by definition, $$T(N)=O(h(N))$$ and $$T(N)=Ω(h(N))$$ must both hold.
checking condition 1, $$2^n*c≥n^2≥n^2-4n$$
and so we choose $$c=5, n=1$$
because $$2^n≥n^2$$ for all$$N≥n=1$$
and we conclude $$T(N)=O(h(N))$$
Now, checking condition 2,
$$c*2^n≤n^2-4n≤n^2 $$
but because we showed that $$2^n≥n^2$$ our check on condition two implies that $$2^n=n^2$$
but if we pick $$c=5$$ we get $$32=25$$ which is untrue and in conclusion we have disproved $$n^2−4n = Θ(2^n)$$

To prove $n^2 - 4 n \ne \Omega(2^n)$ you have to prove that no $c > 0, n_0$ works to make $n^2 - 4 n \ge c \cdot 2^n$ whenever $n \ge n_0$.
Assume it works for $c, n_0$. Thus whenever $n \ge n_0$:
$\begin{align*}
   c \cdot 2^n
     &\le n^2 - 4  n \\
   c
     &\le \frac{n^2 - 4 n}{2^n}
\end{align*}$
If this is true, it has to be that:
$\begin{align*}
   \lim_{n \to \infty} \frac{n^2 - 4 n}{2^n}
       &\ge c
\end{align*}$
It is easy to prove (using e.g. l'Hôpital) that the limit is 0, contradiction.
