Let $f\colon \mathbb R^2 \to \mathbb R^2$ differentiable. Given $Df(0,0)$, show $\lim\limits_{(x,y)\to (0,0)} \frac{||f(x,y)-f(0,0)||}{||(x,y)||}=3$ Let $f\colon \mathbb R^2 \to \mathbb R^2$ differentiable. Given 
$Df(0,0)=
\begin{bmatrix}
0 & 3 \\
3 & 0
\end{bmatrix}$, show that $\lim\limits_{(x,y)\to (0,0)} \frac{||f(x,y)-f(0,0)||}{||(x,y)||}=3$. The hint that I was given is that $Df(0,0)$ is a scalar multiple of the orthogonal matrix $\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}$. This is a problem from an old PhD qualifying exam for my school and not a homework problem. 
 A: $f(x,y)=f(0,0)+Df_{(0,0)}(x,y)+O((x,y))$, this implies that:
$lim_{(x,y)\rightarrow (0,0)}{{\|f(x,y)-f(0,0)\|}\over{\|(x,y)\|}}$
$=lim_{(x,y)\rightarrow (0,0)}{{\|Df_{(0,0)}(x,y)+)(x,y)\|}\over{\|(x,y)\|}}$
$=lim_{(x,y)\rightarrow(0,0)}{{\|(3y,3x)+O(x,y)\|}\over{\|(x,y)\|}}$.
$lim_{(x,y)(\rightarrow(0,0)}{{\|(3y,3x)\|}\over{\|(x,y)\|}}-{\|O(x,y)\over\|(x,y)\|}) \leq lim_{(x,y)\rightarrow(0,0)}{{\|(3y,3x)+O(x,y)\|}\over{\|(x,y)\|}} \leq =lim_{(x,y)(\rightarrow(0,0)}{{\|(3y,3x)\|}\over{\|(x,y)\|}}+{\|O(x,y)\|\over\|(x,y)\|})=3$
since ${{\|(3y,3x)\|}\over{\|(x,y)\|}}=3$ and $lim_{(x,y)\rightarrow(0,0}{\|O(x,y)\|\over\|(x,y)\|}=0$.
A: This is basically the accepted answer. I just want to write it in a different way. I'm not interested in points. I'll use $|\,\,|$ in place of $\|\,\,\|.$
We have
$$f(x,y) - f(0,0) = Df_{(0,0)}(x,y) + r(x,y) = (3y,3x) + r(x,y),$$
where $r(x,y)/|(x,y)| \to 0.$ Thus
$$|(3y,3x)|-|r(x,y)| \le |f(x,y) - f(0,0)| \le |(3y,3x)|+|r(x,y)|.$$
Now $|(3y,3x)|= 3|(x,y)|.$ If we now divide through by $|(x,y)|,$ the squeeze principle gives us the desired limit.
