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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?

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    Take $f(x) = \sqrt{|x|}$, then $\lim_{x \to 0} |f(x)| = 0$, but $|\frac{f(x)}{x}| \to \infty$, as $x \to 0$.2012-05-25

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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.