Given that a function is $o(h)$ if
\begin{equation} \lim_{h \to 0} \frac{f(h)}{h}=0\\ \end{equation}
Why is $P(N(h)>0) \neq o(h)$ (where $N(h)$ is a Poisson process) ?
My attempt
\begin{equation} P(N(h)>(k=0)) = 1-P(N(h)=0) = 1 - \frac {e^{-\lambda h}(\lambda h)^k}{k!}\\ \end{equation}
which gives \begin{equation} P(N(h)>(k=0)) = 1 - \frac {e^{-\lambda h}(1)}{1}\\ \end{equation}
and then dividing by h,
\begin{equation} P(N(h)>(k=0)) = \frac {1 -e^{-\lambda h}}{h}\\ \end{equation}
If I were to take $\lim_{h \to 0}$, wouldn't the numerator essentially be $1-1=0$? How should I approach this?