Show that if d(n) is O( f (n)) and e(n) is O(g(n)), then d(n)−e(n) is not necessarily O( f (n)−g(n))
Let $d(n) = 2n$ and $e(n) = n$. Then
$$d(n)-e(n) = n.$$
Since $d(n) = O(n)$ and $e(n) = O(n)$, we have that $$f(n) = g(n) = n,$$
so
$$O(f(n)-g(n)) = O(n-n) = O(1),$$
and clearly $$n \neq O(1).$$