Prove that $2^n$ is $\Omega(2^{n/2})$

Question

To prove growth rate that $2^n$ is $\Omega(2^{n/2})$ or $O(2^{n/2})$, I can take limit:

\begin{align*} \lim_{n\to \infty} (\frac{f(n)=2^n}{g(n)=2^{\frac{n}{2}}})\\ \lim_{n\to \infty} (\frac{1}{2^{-n} \times 2^{\frac{n}{2}}})\\ = \lim_{n\to \infty} (2^{\frac{n}{2}}) = \infty \end{align*}

Meaning that $f(n) \ge g(n)$ or $f(n) \in \Omega({2^{n/2}})$, using transpose property:

$$f(n) \in \Omega({g(n)}) \iff g(n) \in O({f(n)})$$

We can show that $2^n \ne O(2^{n/2})$.

Question: why $2^n \ne \Theta(2^{n/2})$? Is it because upper bound does not exist please, where $\Theta$ is defined as $2^n = \Theta(2^{n/2}) \iff c_1 2^{n/2} \le2^n \le c_2 2^{n/2}$.

Answer 1

$c_1 2^{n/2} \le2^n \le c_2 2^{n/2}$ is equivalent to $c_1 \le2^{n/2} \le c_2$. But $2^{n/2}$ is not bounded from above.