The $L_p$ norm $\|e^{-x/\epsilon}\|_p=(\frac{\epsilon}{p})^{1/p}(1-e^{-p/\epsilon})^{1/p}$, $1\leq p<\infty$ on the interval $(0,1)$. I am not sure why this norm is $O(\epsilon^{1/p})$ as $\epsilon$ approaches 0. Is it because as $\epsilon$ approaches 0, $e^{-p/\epsilon}$ vanishes leaving $(\frac{\epsilon}{p})^{1/p}$?
Also why is "$e^{-x/\epsilon}$ and zero function become indistinguishable with respect to the $L_p$ norm as $\epsilon$ approaches 0". Because if I take $\epsilon=0.01$ and $p=100$, $\epsilon^{1/p}\approx1$.