# limit with two variables [duplicate]

The limit I need to calculate is $$\lim_{(x,y)\rightarrow (0,0)}\frac{xy^{2}}{x^{4}+y^{2}}$$. Using polar coordinates I get: $$lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^{2}(\theta)}{r^{2}\cos^{4}(\theta)+\sin^{2}(\theta)}$$. Now if $$\sin(\theta)\neq 0$$ then the limit is $$0$$. How do I handle the case where $$\theta=0$$ or $$\theta = \pi$$? And is there a better way to approach this limit?

• When $\sin(\theta )=0$, then obviously $\frac{r\cos(\theta )\sin^2(\theta )}{r^2\cos^4(\theta )+\sin^2(\theta )}=0$. – Surb Oct 31 '20 at 15:47
• Maybe this can help. – user Oct 31 '20 at 15:55

For any $$x\ne 0$$ and $$y\ne 0$$, we have
$$x^4+y^2>y^2>0$$
$$\frac{1}{x^4+y^2}<\frac{1}{y^2}$$
$$|\frac{xy^2}{x^4+y^2}|<|x|$$
By polar coordinates, you will find that $$|F(r,\theta)|\le r|\cos(\theta)|\le r$$