Is $f(n) = \Theta(f(n/2))$?
If I take $f = 2^n$, then whatever value I choose for the constant c, $$2^n \geq c \cdot {2^{n/2}}$$
there exists a value of n, such that $2^n$ is bigger.
How can I show that there is no limit on how large $2^{n/2}$ can be, so that $2^n \neq \Theta(2^{n/2}) $?
You can proceed as follows: for $f$ defined by $f(n)=2^{n}$, we have $f(n/2) = \sqrt{f(n)}$ for every $n\geq 0$, and therefore $$ \frac{f(n)}{f(n/2)} = \sqrt{f(n)} \xrightarrow[n\to\infty]{} \infty $$ However, $f(n) = O(f(n/2))$ would require the ratio to be bounded, namely $$ \lim\sup_{n\to\infty} \frac{f(n)}{f(n/2)} < \infty\,. $$