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Suppose $\lim\limits_{h\to 0}\frac{r(h)}{h} = 0$, does it imply $\lim\limits_{h\to 0} r(h) = 0$?

My reasoning says that the answer is yes.

So if $\frac{r(h)}{h} = 0$, then nothing to prove. If $\frac{r(h)}{h} = a(h)$ where $\lim\limits_{h\to 0} a(h) = 0$, then $r(h) = a(h)h$, hence $$\lim_{h\to 0} r(h) = \lim_{h\to 0} [a(h)h] = \lim_{h\to 0} [a(h)] \cdot \lim_{h\to 0} h$$ (since $\lim ab=\lim a \cdot \lim b$).

And therefore $\lim\limits_{h\to 0} r(h) = 0$.

But if this is true, then why in the definition of differentiation we write

$$\lim_{h\to 0} \frac{|f(x+h) - f(x) - Df(a)(h)|}{|h|} = 0$$ instead of $$\lim_{h\to 0} |f(x+h) - f(x) - Df(a)(h)| = 0?$$

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    Your implication is true, but not the one the other way around. That's the one you need. – 2017-01-10
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    would you elaborate please – 2017-01-10
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    the other way implication is r(h) = 0 which would mean lim r(h)/h = lim 0/h = 0 and hence this becomes if and only if – 2017-01-10
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    No, it becomes $0/0$ (note that the limit takes $h$ to $0$), which is an "indeterminate form" -- that's the whole reason the limit is there. – 2017-01-10

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As mentioned above $\lim_{h \rightarrow 0} r(h) = 0$ does not imply that $\lim_{h \rightarrow 0} \frac{r(h)}{h} = 0$, which is the entire point of the derivative. However the other implication is true, as you correctly identified.

For the derivative to even exist, one necessary condition (but not sufficient) is for $\lim_{h \rightarrow 0} [f(x+h) - f(x)] = 0$ (i.e., continuity), but this certainly does not imply that the derivative of every function $f$ is zero everywhere.