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

Question

Prove $\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2+y}=0$ using epsilon-delta proof.

I tried to solve this problem with epsilon-delta method, but couldn't handle it. anyone help me?

Answer 1

What you're trying to prove is not true. Note that for $x_n=\frac{1}{n}$, $y=-\frac{1}{n^2}+\frac{1}{n^6}$: $$ \frac{xy^2}{x^2+y} = \frac{{1\over n} (-\frac{1}{n^2}+\frac{1}{n^6})^2}{\frac{1}{n^6}} = n(1-\frac{1}{n^4})^2 \rightarrow \infty$$

However for slightly modified function, for example $\frac{xy^2}{x^2+|y|}$, it can be proved. Note that $$|x|,|y|\le \sqrt{x^2+y^2}$$ but for $|y|<1$, $|y|\ge |y|^2$, so $$ x^2+|y| \ge x^2+y^2$$ and $$ \frac{xy^2}{x^2+|y|} \le \frac{(x^2+y^2)^\frac32}{x^2+y^2} = \sqrt{x^2+y^2} \rightarrow 0$$

Answer 2

Note that with $x_n=\frac 1n$, $y_n=\frac1{n^2}-\frac1n$, we have $(x_n,y_n)\to (0,0)$, but $$\frac{x_ny_n}{x_n+y_n}=\frac{\frac1{n 3}-\frac1{n^2}}{\frac1{n²}}=\frac1n-1\to-1$$

Answer 3

Looking at the denominator of $\frac{xy^2}{x^2+y}$ it makes sense to consider a path of approaching $(0,0)$ close to the curve $x^2+y=0$.

So, consider for $y< 0$ and $\alpha > 0$ the path ($y\rightarrow0^-$)

$$(\sqrt{(-y)+(-y)^\alpha},y)\Rightarrow \frac{xy^2}{x^2+y}=\frac{\sqrt{(-y)+(-y)^\alpha}y^2}{(-y)^{\alpha}}\geq \frac{(-y)^{2+\frac{\alpha}{2}}}{(-y)^{\alpha}} = \frac{1}{(-y)^{\frac{1}{2}\alpha-2}}$$

For $\frac{1}{2}\alpha-2 >0 \Leftrightarrow \alpha > 4$ we get $$\frac{1}{(-y)^{\frac{1}{2}\alpha-2}}\stackrel{y \to 0^-}{\longrightarrow}+\infty$$

So, approaching $(0,0)$ along such a path shows that the limit does not exist.