I need to solve the following exercise:
Exercise: What values of $\alpha$, if any, yield $f = \mathcal{O}(\epsilon^\alpha)$ as $\epsilon \downarrow 0$ if $$ f = \dfrac{1}{1 - e^\epsilon} $$
What I've tried:
I know that I can use the following theorem:
If $$ \lim_{e\downarrow e_o}\dfrac{f(\epsilon)}{\phi(\epsilon)} = L, $$ where $-\infty < L < \infty$, then $f = \mathcal{O}(\phi)$ as $\epsilon \downarrow \epsilon_0$.
So I think I need to evaluate the limit $$ \lim_{\epsilon\downarrow 0}\dfrac{1}{\epsilon^\alpha(1 - e^\epsilon)} $$ and find out for which values of $\alpha$ it exists. The limit exists if $\epsilon^\alpha(1 - e^\epsilon)$ converges to infinity or any number other than $0$. I know that $1 - e^\epsilon\to 1 - 1 = 0$ for $\epsilon\downarrow 0$. So the limit exists if $\epsilon^\alpha \to \infty$ quicker than $1 - e^\epsilon \to 0$, because in that case $\epsilon^\alpha(1 - e^\epsilon)\to \infty$ and
$$\lim_{\epsilon\to 0}\dfrac{1}{\epsilon^\alpha(1-e^\epsilon)} = 0$$ If $\alpha > 0$ then $\lim_{\epsilon\downarrow 0}\epsilon^\alpha = 0$ so we know that at least $\alpha \leq 0$ is necessary. If $\alpha = 0$ then $\lim_{\epsilon\downarrow 0}\epsilon^\alpha = 1$. If $\alpha < 0$ then $\lim_{\epsilon\downarrow 0}\epsilon^\alpha = \infty$. Hence I think that for all values of $\alpha < 0$ we have that $f = \mathcal{O}(\epsilon^\alpha)$.
Question: If my reasoning about when the limit exists is correct, I think that the weak point in my solution thus far lies in the fact that I don't really show/know whether or not it is true that $\epsilon^\alpha\to \infty$ quicker than $1 - e^\epsilon \to 0$. Is this true? And if so, how can I show that it is?