Prove that a linear map is 0 if its $o(\|h\|)$.

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Let $L: \mathbb{R}^n \to \mathbb{R}^m$ be a linear map such that $L\vec{h} = o(\|\vec{h}\|)$. Show that $L = 0$. Any help on this?

The assumption means that $$ \lim_{h \to 0} \frac{L(\vec{h})}{\|\vec{h}\|} = 0. $$

How does this imply that $L = 0$?

The only idea so far that I had is that since $L$ is linear we have for the derivative $$ L(\vec{a} + \vec{h}) - L(\vec{a}) = L\vec{h} = o(\|\vec{h}\|) $$ But how to proceed?

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Assume that $$\lim_{h\to 0}{Lh\over|h|}=0\ .$$ Given an arbitrary unit vector $e$, put $h:=t\,e$ with $t>0$. Then $$Le={1\over t}L(t\, e)={Lh\over|h|}\to0\qquad(t\to0+)\ .$$ It follows that the constant $Le$ is in fact $=0$.