# Multivariable limit using polar coordinates?

$$\lim_{(x,y)\to(0,0)} \frac {2\sin (x^2 + y^2) + y^3}{3x^2+3y^2}$$

I am actually very new to polar coordinates so I'm a little bit confused about this part. I substitute this into this: $$\frac {2sin (r^2) + (r cos \theta)^3}{3r^2}$$ But I am not even sure if it's right and what should I do after this. Does the limit even exist? I need an explanation. Thank you!

• Usually one takes $x = r \cos\theta$ and $y = r \sin\theta$. It's possible to take it the other way around, but that will confuse people. Except for that, you're correct so far. Now split the expression into two terms. Notice that $\sin(r^2)/r^2$ is a standard limit. For the second term, simplify it and notice that you get $r$ multiplied with something bounded. Its limit will therefore be $0$. – md2perpe May 16 '17 at 20:53