Question on limits in $\; \mathbb R^n $ including norm Let $\;f:\mathbb R \rightarrow \mathbb R^n \;$ a Lipschitz continuous map such that $\; \lim_{x \to x_0} f(x)=0\;$. I want to see the behaviour of the following limit:
$\; \lim_{x \to x_0} \frac{f(x)}{\vert f(x) \vert}\;$ where $\;\vert \;\; \vert \;$ is the Euclidean norm in $\; \mathbb R^n\;$. 
EDIT: My initial purpose was to solve this $\; \lim_{x \to x_0} \frac{f(x)g(x)}{\vert f(x) \vert}\;$ where $\;g:\mathbb R \rightarrow \mathbb R^n\;$ such that $\; \lim_{x \to x_0} g(x)=0\;$
If $\; f\;$ was a function in $\; \mathbb R\;$ then I could write 
$\; \lim_{x \to x_0} \frac{f(x)}{\vert f(x) \vert}=\begin{cases} 
                                                   \frac{f(x)}{f(x)} & \text{if 
                                                    $f \ge 0$} \\ \frac{f(x)}{-
                                                    f(x)} & \text{if $f \lt 0$} 
                                                   \end{cases}\;$
since the norm would be the absolute value of $\;f\;$ and so the limit wouldn't exist. 
My question : Can I say something similar to the above assuming $\;f \in \mathbb R^n\;$? How should I handle this $\; \lim_{x \to x_0} \frac{f(x)g(x)}{\vert f(x) \vert}\;$?
I would appreciate any help! Thanks in advance!
 A: This limit must not exist! For example take
\begin{align}
\newcommand{\R}{\mathbb{R}}
f:\mspace{0.3em}
\begin{array}{rcl}
\R &\to &\R^2\\
x &\mapsto& x \left(\begin{array}{cc}\cos x \\ \sin x\end{array}\right)
\end{array}
%
\end{align}
Note that $\|f(x)\|_2=|x|$ and $\frac{x}{|x|} = \operatorname{sgn}(x)$ if $x\neq 0$.
Now $\lim_{x\to 0} f(x) = 0$ but
\begin{align}
\lim_{x\to 0+}\frac{f(x)}{\|f(x)\|_2} &=
\lim_{x\to 0+}\left(\begin{array}{cc}\cos x \\ \sin x\end{array}\right) =
\left(\begin{array}{cc}1 \\ 0\end{array}\right) \\
\lim_{x\to 0-}\frac{f(x)}{\|f(x)\|_2} &=
\lim_{x\to 0-}-\left(\begin{array}{cc}\cos x \\ \sin x\end{array}\right) =
\left(\begin{array}{cc}-1 \\ 0\end{array}\right)
\end{align}
So we have that $\lim_{x\to 0+}\frac{f(x)}{\|f(x)\|_2}\neq \lim_{x\to 0-}\frac{f(x)}{\|f(x)\|_2}$ which implies that $\lim_{x\to 0}\frac{f(x)}{\|f(x)\|_2}$ does not exist.
For your initial purpose this doesn't matter because $\frac{f(x)}{\|f(x)\|_2}$ is bounded and $g(x)$ converges to $0$, so the product converges also to $0$. However maybe you want either $f:\R\to\R$, $g:\R\to\R$ or your product is the scalarproduct of $\R^n$. Anyway this will converge to $0$.
