Can I solve $\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2}$ by converting to polar coordinates?

Question

Is it correct to solve this problem like this? $$\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2} $$ $$\lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}$$ $$\frac{x^2+y^2}{x^2y^2}=\lim_{r\to\ 0} \frac{1}{r^2\cos^2\theta\sin^2\theta}=\infty\implies \lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}=0$$

Answer 1

Directly by polar we obtain

$$\frac{x^2y^2}{x^2+x^2y^2+y^2}=\frac{r^2\cos^2\theta \sin^2\theta}{1+ r^2\cos^2\theta \sin^2\theta}\to \frac{0}{1+0}=0$$

Answer 2

You don't have to make it more complicated. Simply observe that $x^2 \le x^2+x^2y^2+y^2\implies \dfrac{x^2y^2}{x^2+x^2y^2+y^2} \le y^2\implies \text{limit} = 0$ .