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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:

$$ \ln(1) = 0$$

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  • $\begingroup$ 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 $(x,y)\to(0,0)$? $\endgroup$
    – A.Γ.
    Oct 5, 2017 at 6:55

1 Answer 1

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Important facts:

  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 $\lim_{(x,y)\to(0,0)}$.

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  • $\begingroup$ In polar coordinates we have; $2{r^2}\ln(r)$ approaches $0$ as $r$ tends to $0$. $\endgroup$
    – M.R. Yegan
    Oct 5, 2017 at 9:14
  • $\begingroup$ This part confuses me why does the limit change to r tends to zero. What if (x,y) approached say (1,5), would r still tend to zero ? Also I can apply the limit to get 2(0)^2 ln(0) = 0 correct ? $\endgroup$
    – Doug Ray
    Oct 5, 2017 at 17:11
  • $\begingroup$ 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 $(x,y)=(0,0)$. $\endgroup$
    – AlgorithmsX
    Oct 6, 2017 at 3:22

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