Continuity of $y \sin(\frac{1}{x})$ Problem: for which values of $\alpha \in \mathbb{R}$ is the function
\begin{equation*}
f: \mathbb{R}^2\to\mathbb{R}^2 \qquad \begin{pmatrix} x \\ y\end{pmatrix} \mapsto \begin{cases} y \sin(\frac{1}{x}) \quad &\text{if $x\neq 0$}\\0  \quad &\text{if $x= 0$}\end{cases}
\end{equation*}
continuous at $a = \begin{pmatrix} 0 \\ \alpha\end{pmatrix}$? Prove your answer.
Attempted solution: I suppose the answer is: $\left\{ \begin{pmatrix} 0 \\ 0\end{pmatrix} \right\}$. I tried to reason as follows, but I am not sure if this is technically correct.
$f$ is continuous at $a$
$\iff \lim_{x\to 0} f\begin{pmatrix} x \\ \alpha\end{pmatrix} = \begin{pmatrix} 0 \\ 0\end{pmatrix}$
$\iff \lim_{x\to 0} \alpha \sin\left(\frac{1}{x} \right)= \begin{pmatrix} 0 \\ 0\end{pmatrix}$
$\iff \alpha = 0$
since $\sin$ is periodic, and thus diverges.
Is this correct?

Your solution looks fine to me.
Notice that $|y\sin{\frac{1}{x}}|=\sqrt{y^2}|\sin{\frac{1}{x}}| \leq \sqrt{x^2+y^2}$
So indeed the function is continuous at $(0,0)$ and only at this point because $\lim_{x \to 0}\sin{\frac{1}{x}}$ does not exist.
