Definition of limit (2 variables)

Question

I have:

$$ f(x,y) = \begin{cases} \dfrac{3x^3y^3}{{x^2+y^4}} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \\ \end{cases} $$

How can I demonstrate the existence of such a limit using epsilon and delta?

I get: $\delta$ = $\frac{\sqrt[4]{\epsilon}}{3}$Definition of limit with 2 variables

Answer 1

It seems you are correct.

Let $\varepsilon > 0$. Choose $\delta = \frac{\sqrt[4]{\varepsilon}}{3}$.

For $\|(x,y)\|_2 < \delta$ we have:

$$\left|\frac{3x^3y^3}{x^2+y^4}\right| = 3 \cdot \underbrace{\frac{x^2}{x^2+y^4}}_{\le 1} \cdot \underbrace{|x|}_{<\delta}\cdot\underbrace{|y|^3}_{<\delta^3} < 3\delta^4 = \varepsilon$$

Therefore

$$\lim_{(x,y)\to(0,0)} \frac{3x^3y^3}{x^2+y^4} = 0$$

so $f$ is continuous at $(0,0)$.