Why does the polar coordinate method not work?

Question

I tried to calculate the limit

$\lim\limits_{(x,y)\rightarrow(0,0)} (x^2+y^2)^x$

By using polar coordinates

$ x = r \cdot \cos(\theta)$

$y = r \cdot \sin(\theta)$

resulting in

$((r\cdot \cos(\theta))^2+(r\cdot \sin(\theta))^2)^{r\cdot \cos(\theta)}$

$ = (r^2(\cos^2(\theta)+\sin^2(\theta))^{r\cdot\cos(\theta)} = (r^2)^{r\cdot\cos(\theta)} = r^{2r\cdot \cos(\theta)}$

which for $r \to 0$ is $1$

But the actual limit of $\lim\limits_{(x,y)\rightarrow(0,0)} (x^2+y^2)^x$ is $0$.

Why does the polar cooridinate method not work?

Thank you in advance.

Answer 1

Consider the logarithm: $$ \lim_{(x,y)\to(0,0)}x\log(x^2+y^2)= \lim_{r\to0^+}2r\cos\theta\log r =0 $$ because $\lim_{r\to0^+}r\log r=0$ and $\cos\theta$ is bounded.

Thus your given limit is $e^0=1$.