$$ \begin{cases}xy^2\sin\left(\frac 1 y\right), \text {if $y$ $\neq$ 0} \\ 0, \text{if $y$ = 0} \end{cases}$$
I'm not sure how to work out this limit as $x$ goes to infinity. If it were just the y part, I know that $y^2$ goes to 0 and $\sin$ is a limited function, so it has to go to 0, but I don't know what to do with the $x$ term.
Best Answer
Let $$ f(x,y)=\begin{cases}xy^2\sin\left(\frac 1 y\right), \text {if $y$ $\neq$ 0} \\ 0, \text{if $y$ = 0} \end{cases}$$
Case 1: Supose $y=0$. Then, $f(x,y) =0$ by definition.
Case 2: Suppose $y\neq 0$. Then, $f(x,y)=xy^2\sin\left(\frac 1 y\right)$. First observe that
$$\Big|f(x,y)\Big|=\Big|xy^2\sin\left(\frac 1 y\right)\Big|\leq\Big|xy^2\Big|$$
since $\Big|\sin\left(\frac 1 y\right)\Big|\leq 1$ by definition. Next, let both $x$ and $y$ approach zero. It is clear that $xy^2$ will also approach zero.
Therefore, in either case, the limit is $0$ which means that $f(x,y)$ is continuous at the origin.