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.

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 $n$.
