Continuity of $x\sin\frac{1}{y}$ at $(x, 0)$ I need to check if function is continue in $(x,0)$
$$f(x,y) =
\left\{
 \begin{array}{ll}
  x\sin\frac{1}{y}  & \mbox{if } y \ne 0 \\
  0 & \mbox{if } y = 0
 \end{array}
\right.$$
Can someone help me understand if I approached this correctly?

First, I check the following limit 
$$\lim_{(x,y) \to (x,0)} x\sin\frac{1}{y}$$
Which does not exist. I can write it as 
$$\lim_{(x,y) \to (x,0)} x \lim_{(x,y) \to (x,0)} \sin\frac{1}{y}$$
The first limit tends to $x$ itself, the second one diverges. 
For $x=0$ the limit 
$$\lim_{(x,y) \to (0,0)} x\sin\frac{1}{y} = 0$$
Indeed, by the squeeze theorem I can write
$$-1\leq\sin\frac{1}{y} \leq 1 $$
$$-x \leq x\sin\frac{1}{y} \leq x$$
Thus reducing the limit to 
$$ \lim_{(x,y) \to (0,0)} x = 0$$
Thus $f(x,y)$ is continuous  $\{ \forall (x, y) \in \Re^2 \mid x \ne 0 \}$
I'm a beginner, thank you in advance! 

It's easier to use sequences: Fix $0\neq x\in \mathbb R$. There are sequences $(y_n)$ and $(z_n)$ such that $y_n\to 0$ and $z_n\to 0$ and such that $\sin(1/y_n)=1$ and $\sin(1/z_n)=-1$ (why?). Then, $\alpha_n=(x,y_n)\to (x,0)$ and $\beta_n=(x,z_n)\to (x,0)$ but $f(\alpha_n)\to x$ and $f(\beta_n)\to -x$ so $f$ is not continuous at $(x,0).$
The case $x=0$ is even easier. 
