Using epsilon delta to show limit for function of two variables

Question

Using epsilon delta I have to show that $$\frac{(e^x-1)y^2}{x^2+y^2} \longrightarrow 0 \ \ \ \ \ \ \text{for} \ \ \ \ \ \ (x,y) \rightarrow (0,0)$$

I don't have problems with one variable but two variables involving makes it difficult. Can someone help?

Answer 1

Hint

$$\left|\frac{y^2(e^x-1)}{x^2+y^2}\right|\leq |e^x-1|.$$

Answer 2

The distance here is

$$d ((a,b),(c,d))=\max (|a-c|,|b-d|) $$

as we are near $(0,0) $. we can assume

$|x|<1$ and $|y|<1.$

also by MVT, we have $$|e^x-1|=|xe^\xi|<e|x|$$ thus

$$|\frac {y^2 (e^x-1)}{x^2+y^2}|<e|x|$$

we can take $$\delta=\frac {\epsilon}{e} . $$