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

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!

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  • $\begingroup$ 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 $0$. $\endgroup$
    – md2perpe
    Commented May 16, 2017 at 20:53

1 Answer 1

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\begin{align} \lim_{(x,y)\rightarrow (0,0)} \frac{2\sin(x^2+y^2)+y^3}{3x^2+3y^2} &= \lim_{r\rightarrow 0} \frac{2\sin(r^2)+r^3\sin^3\theta}{3r^2}\\ &=\lim_{r\rightarrow 0} \frac{2\sin(r^2)}{3r^2} + \lim_{r\rightarrow 0}\frac{r^3\sin^3\theta}{3r^2}\\ &= \frac{2}{3} + \lim_{r\rightarrow 0}\frac{r\sin^3\theta}{3}\\ &= \frac{2}{3} + 0\\ &=\frac{2}{3} \end{align}

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