Study the continuity of a function

Question

I have the function $f:\mathbb{R}^2\rightarrow\mathbb{R}\ f(x,y) = \left\{\begin{matrix} \sin\frac{x^3y}{x^4+y^4}, & (x,y) \in \mathbb{R}^2 \setminus\{(0,0)\}\\ 0, & (x,y) = (0,0). \end{matrix}\right.$
I need to study the continuity of the function $f$.

I tried to calculate the limits as $(x,y) \rightarrow (0,0)$ of $f$. So $\lim_{(x,y)\rightarrow(0,0)}\sin\frac{x^3y}{x^4+y^4} = \sin(0) = 0$ So this would imply that $f$ is continuous at $(0,0)$, right?

Answer 1

HINT:

Note that along $x=0$, $\sin\left(\frac{x^3y}{x^4+y^4}\right)=0$. Along $x=y$, $\sin\left(\frac{x^3y}{x^4+y^4}\right)=\sin(1/2)$. What can you can conclude?

Answer 2

I don't understand your solution. I'll give a hint on how to solve it. What limit do you get if you take $(x,y)\to (0,0)$ along the curve $(t,0)$? And how about the curve $(t,t)$?

Answer 3

$h(x,y)=\frac{x^3y}{x^4+y^4}$ has no limit at $(0;0)$. Puting $x=r\cos{\theta}$ and $y=r\sin{\theta}$ yields: $f(x,y)=\frac{\sin{\theta}\cos^3{\theta}}{\cos^4{\theta}+\sin^4{\theta}}$, choosing $\theta=0$ and $\theta=\frac{\pi}{3}$ provide two different values.