0
$\begingroup$

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:

$$ \ln(1) = 0$$

$\endgroup$
1
  • $\begingroup$ 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

1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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 $(x,y)=(0,0)$. $\endgroup$ Oct 6, 2017 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.