Confirmation of polar coordinate calculation

Question

$$\lim_{x,y \to 0,0} \frac{x^2y + xy^2}{ x^2+y^2}$$

Polar coordinates

$$x= r \cdot \cos \theta$$ $$y = r \cdot \sin \theta$$

$$\lim_{r \to 0} \frac{(r \cos \theta)(r \sin \theta) + (r \cos \theta) (r \cos \theta)}{(r \cos \theta)^2 + (r \sin \theta)}^2 =0$$

Am I right?

Answer 1

No, it is not correct. You should have got$$\lim_{r \to 0} \frac{(r \cos \theta)^2(r \sin \theta) + (r \cos \theta) (r \sin \theta)^2}{(r \cos \theta)^2 + (r \sin \theta)^2}\tag1$$instead. And $(1)$ is equal to $0$ because$$\frac{(r \cos \theta)^2(r \sin \theta) + (r \cos \theta) (r \sin \theta)^2}{(r \cos \theta)^2 + (r \sin \theta)^2}=r\bigl((\cos\theta)^2(\sin\theta)+(\cos\theta)(\sin\theta)^2\bigr).$$

Answer 2

Your algebra is wrong in several places, but the limit is $0$. For an easy solution, squeeze the function between $\pm2r$.

Answer 3

Polar is fine, but in this particular case, this below is way simpler:

$$|x^2y+xy^2|\le |y|x^2+|x|y^2\le \underbrace{\max(|x|,|y|)}_{\to 0}\ (x^2+y^2)$$