what does the "element of notation" means in big O statments? what does the following means?
$2^{\sqrt{x}} \in O(2^x)$
I have given the above statement to determine whether it is true or not but I do know the meaning of the $\in$ symbol
Notation $O(f),x \to x_0$ is set of functions:
$$O(f(x))=\{g: \exists U(x_0) \text{ set},\exists C&gt;0, \forall x \in U(x_0), |f(x)| \leqslant C|g(x)|  \}  $$
so, $g \in O(f)$ mean that it satisfies above condition.
Also from above is clear, that without pointing $x_0$ set $O(f)$ have no sense.
For your example: if we take $x_0=0$, then your statement is not true, because for $x \in (0,1)$ we have $2^{\sqrt{x}}&gt;2^x$. If we take any $x_0&gt;1$, including $x_0=\infty$, then reverse is true for some $U(x_0)$ neigbourhood of $x_0$ and your statement is true. So, you cannot answer without $x_0$.
Hint: To show that $2^{\sqrt{x}} \in O(2^x)$, it suffices to show that
$$
\lim_{x \to \infty}\frac{2^{\sqrt{x}}}{2^x} = 0.
$$