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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 $$ \frac{n}{2}=\Theta(n) $$

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Yes (trivally). See… – M.B. Jan 24 '12 at 9:39
In general, for any function $f(n)$ and any constant $c$, we have $c f(n) = \Theta(f(n))$. – Srivatsan Jan 26 '12 at 16:01
up vote 3 down vote accepted

Yes but note that it also immediately follows from the definitions.

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you mean definition of $\Theta(n)$? – user19821 Jan 24 '12 at 10:12
@user19821 yes indeed, it is obvious how you have to choose the constants that it nicely fits the definition. – Listing Jan 24 '12 at 22:31

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