Finding the value of c for a two variable function to allow continuity.

Question

The function f:$\mathbb{R^2}$$\to$$\mathbb{R}$ given by $$f(x,y)=\begin{cases}\frac{xy^2}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ c & \text{if }(x, y) = 0\end{cases}.$$

I am trying to calculate the limit (by any means) to determine the value of constant $c$ such that $f$ is continuous at $(0,0)$.

I have obtained the limit for $$f(x,y)=\frac{xy^2}{x^2+y^2} \quad \text{if } (x,y) \neq (0,0)$$ through using polar coordinates $L=0$.

I do not know how to find $c$ from here.

Answer 1

Observe that $$\left | \frac {xy^2} {x^2+y^2} \right | \leq \sqrt {x^2+y^2} < \varepsilon$$ whenever $\sqrt {x^2 + y^2} < \delta,$ where $\delta = \varepsilon.$

So $$\lim\limits_{(x,y) \rightarrow (0,0)} \frac {xy^2} {x^2 + y^2} = 0.$$

For the continuity of $f$ at $(0,0)$ we should have $c=f(0,0) = \lim\limits_{(x,y) \rightarrow (0,0)} \frac {xy^2} {x^2 + y^2} = 0.$

Answer 2

Hint: Use that $$\frac{x^2+y^2}{2}\geq |xy|$$

Answer 3

Do you that something that converges to zero times something bounded converges to zero? Well:

$$\frac{|xy^2|}{x^2+y^2}=|x|\frac{y^2}{x^2+y^2}\le|x|\frac{x^2+y^2}{x^2+y^2}\ldots$$