Let $m > 0$ and $ a:[0,1] \rightarrow \mathbb R$ be a function with $a(\epsilon) \rightarrow 0$.
Then $ \epsilon^m a(\epsilon) \rightarrow 0 $ for $ \epsilon \rightarrow 0$.
But is $ \epsilon^m a(\epsilon) \in o(\epsilon^m) $, with O-Notation used? Put into other words, if we multiply a zero sequence which converges by order $m$ with another zero sequence, is the decay strictly faster than by order $m$?
You may assume $a$ to be uniformly continuous if that helps.