# Asymptotic behavior of $1/\ln n$ and $e^x-1$

I have two asymptotic notation questions.

Determine whether or not $\frac{1}{\ln(n)}=O(\frac{1}{n})$

My answer I put the statement is not true. A logarithmic function $\frac{1}{\ln(n)}$ grows faster $\frac{1}{n}$ regardless of constant placed on $\frac{1}{n}$.

Thus $\frac{1}{\ln(n)}$ not upper bound by $O(\frac{1}{n})$

My second question is

Show that $e^x-1=O(x^2)$ is not true.

But I think it is true because using a graph you see

using a constant c=3 $e^x-1=O(3x^2)$ for $x>0.413$ but I am not sure if this is correct.

• No, it is not correct. $e^x-1$ grows much faster, because it is "an exponential". . "Using a graph" is not really convincing here. – Dietrich Burde Sep 6 '15 at 18:38
• Yes I think because e^x-1 is exponetial. – Fernando Martinez Sep 6 '15 at 18:41
• What is the definition of big O that you are using? Normally, this is defined with the limit of a ratio, and this should be a useful way to proceed in both questions. – process91 Sep 6 '15 at 19:06
• I have f(x)=0(g(x)) f(x)<=Cg(x) for all x>N – Fernando Martinez Sep 6 '15 at 19:41
• Big O notation is the proper formalization for growing slower/faster than. So you should better use the formal definition, instead of intuitive arguments. In both cases, you could use l'hopital. – user251257 Sep 7 '15 at 1:17

From the power series for $e^x$, $e^x > \frac{x^{n+1}}{(n+1)!}$, so $\frac{e^x}{x^n} >\frac{x}{(n+1)!} \to \infty$, so $e^x \ne O(x^n)$ for any $n$.