Logarithm multivariable limit $\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$

Question

Find multivariable limit $$\lim_{\left( x,y \right) \rightarrow (0,0)}\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$$ I was trying to find and inequality i've found out that: $$\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2} \ln(1+x^3+y^3)}{2xy} $$ and after that i am stuck. From iterated limits i've calculated before i knwo that the limit exists for certain. I have also another idea to do that precisely to $y=x$ and $y=-x$ but i do not know whether this is working properly.

Answer 1

Using Polar coordinates.

Let $x = r\cos(\theta)\ and \ y=r\sin(\theta)$ hence the function becomes $\rightarrow$ $\frac{\ln( 1 +r^3\cos^3(\theta)sin^3(\theta))}{r}$ .

As $r^3\cos^3(\theta)\sin^3(\theta)$ $\rightarrow$ $0$ as $r$ $\rightarrow$ $0$ hence $\ln(1 + r^3\cos^3(\theta)\sin^3(\theta))$ $\sim$ $r^3\cos^3(\theta)\sin^3(\theta)$ and so the function becomes $\frac{r^3\cos^3(\theta)\sin^3(\theta)}{r}$ that tends to $0$ as r$\rightarrow$ $0$.