Is this function $f$ continous? Let $f:\mathbb{R}^2\to\mathbb R$ with 
$$
f(x,y)=\begin{cases}\frac{x^2}{||(x,y)||}&\text{for }(x,y)\neq(0,0)\\ 0&\text{for }(x,y)=(0,0)\end{cases}
$$
The norm is not further specified. As every norm is continuous and $||(x,y)||=0$ iff $(x,y)=(0,0)$, we have that $f$ is continuous for every $(x,y)\neq (0,0)$(as $x^2$ is obviously continuous).
(0,0) gives me more troubles as I don't know how to handle the unspecified norm. I know that all norms over $\mathbb R^2$ are equivalent, so there may be a way to convert this problem into a version using the classical Euclidean norm.
 A: Since the unit circle is compact, any continuous function on the unit circle has a minimum and maximum.  In particular, there are positive numbers $m$ and $M$ so that
$$
m\leq\|(\cos(\theta),\sin(\theta))\|\leq M.
$$
These numbers are positive since a norm is zero if and only if its input is zero.
Suppose that $\|(x,y)\|_2=\lambda$, where the subscript of $2$ represents the Euclidean norm.  Then, we know that $-\lambda\leq x\leq \lambda$ and that $(x,y)=\lambda(\cos(\theta),\sin(\theta))$ for some $\theta$.  Therefore,
$$
0\leq \frac{x^2}{\|(x,y)\|}\leq \frac{\lambda^2}{\lambda m}=\frac{\lambda}{m}.
$$
Since all norms induce the same topology on $\mathbb{R}^2$, regardless of the norm, if $(x,y)\rightarrow(0,0)$, then $\|(x,y)\|_2,\|(x,y)\|\rightarrow 0$.  Therefore, we may apply the squeeze lemma and get that the limit exists. 
A: To finish the question, it suffices to prove the continuity of $f$ at the origin.
Since $\Bbb{R}^2$ is finite-dimensional, all norms on $\Bbb{R}^2$ are equivalent,
$$\text{i.e. } \exists c,C \ge 0 \text{ s.t. } c\sqrt{x^2+y^2}\leqslant \|(x,y)\|\leqslant C\sqrt{x^2+y^2}$$
so it suffices to prove this using the Euclidean norm.
Let $\epsilon > 0$.  Take $\delta = \epsilon$.  For all $||(x,y)|| < \delta$,
$$|f(x,y) - f(0,0)| = \left|\frac{x^2}{||(x,y)||} - 0 \right| < \frac{x^2}{\sqrt{x^2+y^2}} \le \frac{x^2+y^2}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2} \\ < \frac1c ||(x,y)|| = \frac{\delta}{c} = \frac{\epsilon}{c}.$$
Hence $f$ is continuous at $(0,0)$.
A: Hint: any two norms on a finite dimensional vector space are equivalent.
Therefore there are positive constants $c$, $C$ such that for all $x$, $y$,$$c\sqrt{x^2+y^2}\leqslant \|(x,y)\|\leqslant C\sqrt{x^2+y^2}.$$
