Answer
if $f(x)$ is $O(g(x))$ and $g(x)$ is $O(h(x))$
then $f(x)$ is $O(h(x)$
Work Step by Step
Given: $f(x)$ is $O(g(x))$ and $g(x)$ is $O(h(x))$
We have to prove that $f(x)$ is $O(h(x))$
Since $f(x)$ is $O(g(x))$, there exist constants $C_{1}$ and $k_{1}$ such that:
$|f(x)| \leq C_{1}|g(x)|$
When $x>k_{1}$
Since $g(x)$ is $ O(h(x))$, there exist constants $C{2}$ and $k_{2}$ such that:
$|g(x)| \leq C_{2}|h(x)|$
When $x>k_{2}$.
For convenience sake, we choose $k={k_2}$ (which also implies $x>k_{2}$) and $C=C_{1} C_{2}$.
$|f(x)| \leq C_{1}|g(x)| \leq C_{1}\left(C_{2}|h(x)|\right)=C_{1} C_{2}|h(x)|=C|h(x)|$
By the Big-O definition, we then obtained that $f(x)$ is $O(h(x))$ with constants
$k=k_{2}$ and $C=C_{1} C_{2}$.
