So I'm not sure how much background I need to give to set up this question. But in my lecture notes I have that
$e^{-\eta / \epsilon^{1-\alpha}}$
can be ignored where $\epsilon << 1$ and $0 <\alpha <1$ and $\eta$ is fixed. In particular that:
$\lim_{\epsilon \rightarrow 0} exp(-\eta/ \epsilon^{1-\alpha})/\epsilon^k = 0$ for all $k \ge 0$
I tried playing around with this limit with L'Hopitals but I wasn't really sure how to show it rigorously. Applying L'Hopitals it seemed like the limit is equal to
$\lim_{\epsilon \rightarrow 0} \eta(1-\alpha)exp(-\eta/ \epsilon^{1-\alpha})/k\epsilon^{k-1 + (1-\alpha)^2} \le \lim_{\epsilon \rightarrow 0} \eta(1-\alpha)exp(-\eta/ \epsilon^{1-\alpha})/k\epsilon^{k-1}$
So then it seems like when you keep applying L'hopitals, the constants change but you will keep getting the power of the denominator to decrease by 1, so it seems like it will go to zero. But I'm not sure how to make this rigorous. Is this argument even the right track? How would you show this limit? Thanks for your time!
One way is to take the logarithm: $$-\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon$$ Here $\log\epsilon$ is much smaller than the first term, so the expression tends to $-\infty$. One way to show it from scratch is to consider $$f(\epsilon)=-\frac12\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon$$ observe that $f'(\epsilon)>0$ for all small $\epsilon$, and conclude that $f$ is bounded from above in a neighborhood of zero. Say: $f\le M$, then $$-\frac{\eta}{\epsilon^{1-\alpha}} - k\log \epsilon\le -\frac12\frac{\eta}{\epsilon^{1-\alpha}} +M $$ which tends to $-\infty$ as $\epsilon\to 0$.