# Two variable-limit.

I've trying to find this:

$$\lim_{(x,y)\rightarrow(0,0)} (x+y^2)\sin(\frac{1}{xy})$$ And I've just converted to polar coordinates: $$\lim_{\rho\rightarrow 0} (\rho \cos \theta + \rho^2 \sin^2 \theta)\sin(\frac{1}{\rho^2 \cos \theta \sin \theta}) =$$ $$\lim_{\rho\rightarrow 0} \rho \cos \theta \sin(\frac{1}{\rho^2 \cos \theta \sin \theta}) + \lim_{\rho\rightarrow 0} \rho^2 \sin^2 \theta \sin(\frac{1}{\rho^2 \cos \theta \sin \theta})$$

Then, it follows that $|\rho\cos\theta\sin(\frac{1}{\rho^2 \cos \theta \sin \theta})| \leq |\rho \cos \theta| \leq |\rho|$, so the first limit must converge to zero. Something similar can be said about the second limit: $|\rho ^2 \sin ^2 \theta \sin(\frac{1}{\rho \cos \theta \sin \theta}) \leq |\rho ^2 \sin ^2 \theta| \leq |\rho^2|$, so the second limit must also be equal to zero. Nevertheless, WolframAlpha says the limit does not converge because it's path-dependent. Have I gone wrong somewhere? I don't see a $\theta$ in that limit which may cause any problems.

PS: I'm aware that the $\sin(\frac{1}{\rho^2 \cos \theta \sin \theta})$ will oscillate very wildly as $\rho$ approaches zero, but in the end, the sine function is bounded, and I still don't undertand where's my mistake.

• $sin(\frac{1}{xy})$ is limited by $-1$ and $1$. And you have $\lim_{(x,y)\rightarrow (0,0)}(x+y^{2})=0$. – Mateus Rocha Jul 3 '18 at 15:19

Hint: we have $$|x+y^2||\sin(\frac{1}{xy})|\le |x|+y^2$$ and this tends to Zero for $x,y$ tends to Zero.
• I don't know, the Limit is Zero,this says my proof, the reason is that $$|\sin(\frac{1}{xy})|\le 1$$ – Dr. Sonnhard Graubner Jul 3 '18 at 15:27