I need to show that the function $f(x,y)=\frac{x^5+y^6}{|x|^3+|y|^3}$ is continuous at (0,0), so I want to show that $ \lim_{(x,y)\to(0,0)}\frac{x^5+y^6}{|x|^3+|y|^3} =0$
I tried switching to polar coordinates but did not see how that helps. what would be a good way to calculate the limit?
It's enough to show $\;\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}$ tend to $0$. Now, setting $x=r\cos\theta$, $y=r\sin\theta$, we have $$\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}\le\dfrac{\lvert x\rvert^5+\lvert y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}=\frac{r^5\bigl(\lvert\cos\theta\rvert^5+r\lvert\sin\theta\rvert^6\bigr)}{r^3\bigl(\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3}\le\frac{r^2(1+r)}{\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3}.$$ There remains to show $\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3$ has a positive lower bound. For this, we may suppose $0\le \theta\le \pi/2$, so that \begin{align} \lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3&=\cos\theta^3+\sin^3\theta=(\sin\theta+\cos\theta)(1+\sin\theta\cos\theta)\\&\ge\sin\theta+\cos\theta=\sqrt2\sin\Bigl(\theta+\frac\pi4\Bigr)\ge1\qquad\text{if }\;0\le \theta\le \frac\pi2. \end{align} Thus $\;\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}\le r^2(1+r)$, which tends to $0$ when $r$ tends to $0$.