Help with Multivariable Limits

Currently struggling over a limit problem I've gone back and forth with for a few hours now. Any help would be appreciated.

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

I tried converting to polar coordinates, but the pesky $$\ln (0)$$ doesn't seem to go away. The only thing I can really do is prove it does exist, but I don't suppose that does much for me. Any help would be great.

• Does that not still give you ln(0) afterwards? If you do y = mx, you get sin(x^2 + m^2*x^2)ln(x^2 + m^2*x^2)? Which doesn't crystallize to much? Commented Sep 30, 2017 at 20:08

Take $$x=r\cos{t}$$ $$y=r\sin{t}$$

Then $$|f(r,t)|=|2\sin{r^2}\ln{r}| \leq 2r^2|\ln{r}|\to 0$$ as $r \to 0^+$

This is true from L'Hospitals rule because:

If $0<r<1$ then $\ln{r}<0 \Rightarrow |\ln{r}|=-\ln{r}$ and also:$$\lim_{r \to 0^+}2r^2|\ln{r}|=\lim_{r \to 0^+}\frac{-2\ln{r}}{\frac{1}{r^2}}=^{L.H}\lim_{r \to 0^+}-2r = 0$$

• I miss the jump you make when you remove sin(r^2) and replace it with r^2. Are you using the squeeze theorem after that? Commented Sep 30, 2017 at 20:39
• @Wills23 i used $|\sin{x}| \leq |x|$ and then squeeze theorem Commented Sep 30, 2017 at 20:41