<s>[INST] Continuity of functions of two variables Find the points where the function $f(x,y)$ is continuous where
$$f(x,y)=\left\{ \begin{array}{ll}
\frac{x^2\sin^2y}{x^2+2y^2} & \mbox{if $(x,y)\not=(0,0)$};\\
0 & \mbox{if $(x,y)=(0,0)$}.\end{array} \right.$$  
What I attempted: Here $f(x,y)$ is continuous at all the points $(x,y)\not=(0,0)$  
We will check continuity at the point $(x,y)=(0,0)$  
$f(x,y)=0$ is well defined. Now,  
$$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}$$  
Here $0\le \frac{x^2\sin^2y}{x^2+2y^2}\le \frac{x^2y^2}{x^2+2y^2}$ (As $\sin^2y \le y$)  [Am I correct here?]
If we let  $x=r\cos\theta$, $y=r\sin\theta$, then $$\frac{x^2y^2}{2x^2+y^2}=\frac{r^2 \cos^2\theta r^2 \sin^2\theta}{r^2+r^2 \sin^2\theta}=\frac{r^2 \sin^2\theta \cos^2\theta }{1+\sin^2\theta}\le r^2$$ (As $\frac{\sin^2\theta}{1+\sin^2\theta}\le1$, $\cos^2\theta\le 1$ )  
As $(x,y)\to (0,0)$, $r\to 0$. Therefore using Sandwich Theorem we should have $$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}=0$$
So, it is continuous at $(0,0)$.  
I am not sure whether it is correct or not.The question was asked in an interview and I used the above technique. However, they did not pointed my mistake but just gave an another problem of an similar kind where $f(x,y)=\frac{x^2+\sin^2y}{2x^2+y^2}$ for all $(x,y) \not= (0,0)$ and $0$ for the origin.I tried the same method but failed. Then they asked me to leave.
 [/INST] Here is the answer for the second question: $\frac {x^{2}+\sin ^{2} (y)} {2x^{2}+y^{2}}$ is $\frac  1 2$ when $y=0$ and $\frac {\sin ^{2} (y)} {y^{2}}$ when $x=0$. The last expression approaches 1 as $y \to 0$. Hence the functioin does not even have a limit as $(x,y) \to 0$ and the function is not continuous. 
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