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! 

I would split up the analysis into cases.
Case 1: If $y=0$, then $f(x,y)=0$ by definition.
Case 2: If $x=0$ and $y\neq 0$, then 
$$\Big|f(x,y)\Big|=\lim_{(x,y) \to (0,y)} \Big|x\sin\frac{1}{y}\Big|\leq \lim_{(x,y) \to (0,y)}\Big|x\Big|$$
since $|\sin\frac{1}{y}|\leq 1$. In this case, $f(x,y)\to 0$ since $x=0$.
Case 3: If $x\neq 0$ and $y\neq 0$, then
$$\Big|f(x,y)\Big|=\lim_{(x,y) \to (x,y)} \Big|x\sin\frac{1}{y}\Big|\leq \lim_{(x,y) \to (x,y)}\Big|x\Big|$$
since $|\sin\frac{1}{y}|\leq 1$. However, as $x\neq 0$, $f(x,y)\not \to 0$. Therefore, $f(x,y)$ will be greater or less than zero depending on the value of $x$.
So, the function $f(x,y)$ isn't continuous at $(x,0)$. 
It is true that $f(x,y)$ is continuous at $(x,0)$ provided that $\{(x,y)\in
\mathbb R^2∣(x=0~ \text{and}~ y≠0)~ \text{or}~y=0\}$.
