Asymptotic Notation - Big/little o I know that $H = O(h)$ as $h \to 0$.
And we have a term $T$ which is $o(H)$ as $H \to 0$.
I want to show that this means $T$ is $o(h)$ as $h \to 0$.
I understand intuitively that this means that $T$ grows at a rate which is much smaller than a rate which is a fixed multiple of the rate of $h$ and so the above is true.
I feel that I could prove it if I can get the limits to be in the same variable.
Can somebody provide a detailed proof with explanations of each step taken?
 A: Just unravel the definitions. $T = o(H)$ means for every $\epsilon > 0$ there's a $\delta$ such that $$|T(x)| \le \epsilon|H(x)| \tag{1}$$
for all $|x| < \delta$. And $H = O(x)$ means there's a $C$ and a $\delta_0$ such that $$|H(x)| \le Cx \tag{2}$$ for all $|x| < \delta_0$. 
Combining (1) and (2) gives
$$|T(x)| \le C\epsilon x \tag{3}$$
for all $|x| < \min\{\delta, \delta_0\}$.
This proves $T(x) = o(Cx)$. Convince yourself that $o(Cx) = o(x)$.
A: If you write everything in terms of functions it's not so bad: you have a function $H(h)$ which is $O(h)$ as $h \to 0$ and a function $T(H)$ which is $o(H)$ as $H \to 0$. You now consider $f(h)=T(H(h))$. Now as $h \to 0$, $H(h)$ goes to zero. Thus $f(h) = C_1(h) H(h)$ where $C(h) \to 0$ as $h \to 0$. But now $H(h)=O(h)$ so $f(h) \leq C_2(h) h$, where $C_2(h)$ is $C_1(h)$ times something bounded, so it also goes to zero. Thus $f(h)=o(h)$. 
A: To show $T=o(h)$ is to show $\frac{T}{h}\to 0$ as $h\to 0$.
So let $\epsilon>0$ be given. Because $H=O(h)$, there is $\delta>0$ and $M>0$ such that $|h|<\delta$ implies $|H|\leq M|h|$. Next, because $T=o(H)$, there is $\delta_1$ such that $|H|<\delta_1$ implies $|T/H|<\epsilon/M$. It remains to pick $\delta_2=\min\{\delta,\delta_1/M\}$ and conclude:
$$
|h|<\delta_2\implies|T/H|\times|H|<(\epsilon/M)\times M|h|=\epsilon|h|\implies|T/h|<\epsilon.
$$
