# Limit in two variables with polar coordinates

i have a problem with this exercise.

Given the function $\left\{\begin{matrix} \frac{y^2|xy|(x^3-y^2)}{\sqrt{x^2+y^2}} if (x,y)\neq (0,0) \\ 0 if (x,y)=(0,0)\end{matrix}\right.$ test its continuity.

The function is defined for ${(x,y)\in \mathbb{R}^2:\sqrt{(x^2+y^2)}>0}$, so it's a continuos function in its domain and it's worth zero in $(0,0)$. To test the continuinity in point i pass to the polar coordinates. Now, i'm arrived to proof that:

$\lim_{\rho ->0^+}|\rho ^5sin^2\Theta |cos\Theta |\cdot |sin\Theta |(\rho cos^3\Theta-sin^2\Theta)|$

So:

• for $cos\Theta$ and $sin\Theta$ i can use $|sin\Theta|\leq 1$ and $|cos\Theta|\leq 1$

• for $|(\rho cos^3\Theta-sin^2\Theta)|$ i can use the inequality triangulate: $|(\rho cos^3\Theta-sin^2\Theta)|=|\rho|cos^3\Theta|+|sin^2\Theta||$

but then? How can i eliminate the "first" $|sin^2\Theta|$ (the one next $\rho^5$)? How can i say about $|cos^3\Theta|+|sin^2\Theta|$?

Thanks for any help!

If $x = r\cos\theta$ and $y = r\sin\theta$, then since $|\cos\theta|\leq 1$ and $|\sin\theta|\leq 1$ (so that $|\cos\theta|^3\leq 1$ and $|\sin\theta|^2 \leq 1$), we get \begin{align} \left|\frac{y^2|xy|(x^3-y^2)}{\sqrt{x^2+y^2}}\right| &= \left|\frac{r^6\sin^2\theta|\cos\theta\sin\theta|(r\cos^3\theta-\sin^2\theta)}{r}\right| = r^5 |\sin^3\theta\cos\theta(r\cos^3\theta+\sin^2\theta)|\\ &\leq r^5(r|\cos\theta|^3+|\sin\theta|^2)\leq r^5(r+1) \to 0 \end{align} as $r \to 0$.
• Thanks for the answer. So, in general, we can say that $|sin^n\Theta|<=1$ because $|sin^n\Theta|=|sin\Theta|^n\leq 1$, right? Sep 12, 2018 at 17:46
• @MarcoPittella Yes! Assuming that $\sin^n\theta$ is well defined of course.