This is probably a very silly question.
If $h(n)=\frac{n}{2}, \ g(n)=n$, so $ \lim_{n \to \infty} \frac{h(n)}{g(n)} = \lim_{n \to \infty} \frac{n}{2n}=\frac{1}{2} $ so $h(n) \leq C_1 g(n), h(n)=O(n)$
at the same time , if $g(n)=\frac{n}{10}$, this limit becomes 5, so
$h(n) \geq C_2 g(n), h(n)= \Omega(n)$
Is this logic correct enough to say that $ \frac{n}{2}=\Theta(n) $