Little-o notation for limits at zero

Question

Very quick question which will help me clarify my understanding. Let's say I have an expression that uses little-o notation like the following $$o_{h\to 0}(h)(1-h)+o_{h\to 0}(h)$$ Would I be correct in saying that this can be simplified like this: $$o_{h\to 0}(h)(1-h)+o_{h\to 0}(h)=o_{h\to 0}(h)-o_{h\to 0}(h^2)+o_{h\to 0}(h)=o_{h\to 0}(h)-o_{h\to 0}(h^2)=o_{h\to 0}(h).$$ I suspect that this is true since when we take $h\to 0$, the $o_{h\to 0}(h^2)$ converges faster than $o_{h\to 0}(h)$?

Answer 1

Let $E$ be your expression. You could consider taking the absolute value of $E$ and use the triangle inequality to yield $$\vert E \vert \leq o_{h \to 0}(h) + o_{h \to 0}(h + h^2) \leq o_{h \to 0}(h) + o_{h \to 0}(h^2) \leq \frac{c}{2} (h + h^2) \leq c h$$ for sufficiently small $h,$ where $c \in \mathbb{R}.$ Then of course $\vert E \vert = o_{h \to 0}(h).$ I hope this helps. :)