Are functions that are equal to a constant as you take a limit to infinty O(1)?
By definition of $O$ notation, if you can find a constant $c$ and $n_0$ such that $f(n) &lt; c g(n)$ for all $n &gt; n_0$, we can say $f(n) = O(g(n))$. In this case, suppose $n_0 = 1000$ and $c = 3$ and $g(n) = 1$. You can show that as $f(n)$ is increasing function and $\lim_{n\to\infty}f(n) = e$, it is true $f(n) &lt; 3 g(n)$ for all $n &gt; 1000$. Therefore, $f(n) = O(1)$.