# Solve the limit using polar coordinates

I am given the limit :

$$\lim_{(x,y)\to (0,0)}\left[ \ x^2 + y^2 \ln(x^2+y^2)\right]$$

to solve using polar coordinates first I would convert the equation to polar coordinates : $$\lim_{(x,y)\to (0,0)} \left[\sin^2(\theta) + \cos^2(\theta) \ln(\sin^2(\theta) + \cos^2(\theta)) \right]$$

can I apply substitution to get:

$$\ln(1) = 0$$

• You probably mean $(x^2+y^2)\ln(x^2+y^2)$. How come you substitute $x=\sin\theta$, $y=\cos\theta$ and think that $(x,y)\to(0,0)$?
– A.Γ.
Oct 5, 2017 at 6:55

1. $$x=r\cos\theta$$
2. $$y=r\sin\theta$$
3. \begin{align}x^2+y^2&=r^2\cos^2\theta+r^2\sin^2\theta\\&=r^2(\cos^2\theta+\sin^2\theta)\\&=r^2\end{align}
This means you can substitute $r^2$ in for $x^2+y^2$, $r\cos\theta$ for $x$, $r\sin\theta$ for $y$, and $\lim_{r\to 0}$ in for $\lim_{(x,y)\to(0,0)}$.
• In polar coordinates we have; $2{r^2}\ln(r)$ approaches $0$ as $r$ tends to $0$. Oct 5, 2017 at 9:14
• More specifically, if $(x, y)\to(a,b)$, then $(r,\theta)\to(\sqrt{a^2+b^2},\arctan(\frac yx))$. In your case, $\theta$ is irrelevant, so it can be ignored, and $r=0$ when $(x,y)=(0,0)$. Oct 6, 2017 at 3:22