Is $f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$ uniformly continuous or not Find out if function $$f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$$ is uniformly continous or not in area $D=\{0<x^2+y^2<1\}$.
I found out that we have no $\lim\limits_{x,y\to 0}f(x,y)$, because $$\lim\limits_{x,y\to 0}\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\rho^3\sin^3\alpha+\rho^3\cos^3\alpha}}{\sqrt[5]{\rho^5\sin^5\alpha+\rho^5\cos^5\alpha}}=\lim\limits_{\rho\to0}\frac{\sqrt[3]{\sin^3\alpha+\cos^3\alpha}}{\sqrt[5]{\sin^5\alpha+\cos^5\alpha}}$$
hence we can't use The Uniform Continuity Theorem as we can't determ $f(0,0)$. Function doesn't have bounded partial derivatives, so I think it's not uniformly continous, but I don't know how to show that 

It's not uniformly continuous. Take $x = \rho\cos{\alpha},y= \rho\sin{\alpha}$. Let $\alpha =\pi/4,$, and let $\rho \rightarrow 0^+$, then $f(x,y) \rightarrow \frac{2^{1/3}}{2^{1/5}}$, but take $\alpha =0$, then $f(x,y) \rightarrow 1$.
So let $\epsilon = \frac{1}{2}|\frac{2^{1/3}}{2^{1/5}} - 1|$, then $\forall \delta > 0$, we can find very small $\rho_1, \rho_2>0$, such that $f(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4})$ is very close to $\frac{2^{1/3}}{2^{1/5}}$, and $f(\rho_2 \cos{0},\rho_2\sin{0})$ is very close to $1$, and $(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}), (\rho_2 \cos{0},\rho_2\sin{0})$ are very close to each other. 
Then $||(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}) - (\rho_2 \cos{0},\rho_2\sin{0})|| < \delta$, $|f(\rho_1 \cos{\pi/4},\rho_1\sin{\pi/4}) - f(\rho_2 \cos{0},\rho_2\sin{0})| > \epsilon$. 
Thus $f(x, y)$ is not uniformly continuous in any neighborhood of $(0,0)$.
