$\lim_{x\rightarrow a}\|f(x)\|$ and $\lim_{x\rightarrow a}\frac{\|f(x)\|}{\|x-a\|}$  Given any function $f: \mathbb{R^n} \to \mathbb{R^m}$ , if $$\lim_{x\rightarrow a}\|f(x)\| = 0$$ then does $$\lim_{x\rightarrow a}\frac{\|f(x)\|}{\|x-a\|} = 0 $$
as well?  Is the converse true?

For the first part, consider e.g. the case $m = n$ with $f$ defined by $f(x) = x - a$ for all $x$ to see that the answer is no.
For the second part, the answer is yes.  If $\lim_{x \to a} \|f(x)\|/\|x-a\| = L$ exists (we do not need to assume that it is $0$), then since $\lim_{x \to a} \|x - a\| = 0$ clearly exists, we have that
$$
\lim_{x \to a} \|f(x)\| = \lim_{x \to a}\left( \|x - a\| \cdot \frac{\|f(x)\|}{\|x-a\|}\right) = \lim_{x \to a} \|x - a\| \cdot \lim_{x \to a} \frac{\|f(x)\|}{\|x-a\|} = 0 \cdot L = 0
$$
exists and is $0$ by standard limit laws.
