Show this function of two variables $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at (0,0)

Question

Is the function $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at $(0, 0)$?

Trying to use the epsilon-delta definition of continuity to prove this but can't figure it out.

$\sqrt{x^2+y^2} < \delta$ implies $\left|\frac{2x^2y^3}{x^4+y^6}\right| < \epsilon$.

Using polar coordinates $r < \delta$ implies $$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| < \epsilon^4$$

$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| \leq \left|\frac{2r}{\cos^4(\theta)+r^2\sin^6(\theta)}\right|$$

But from here I cant simplify the equation further.

Answer 1

Note $f(x,y)$ is continuous at $(0,0)$ if $\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$, which is equivalent to for all $(x_n,y_n)$ such that $x_n\to 0,y_n\to 0$, we have $\lim f(x_n,y_n)$ exists and equal to $f(0,0)$.

In your example, the function is not continuous at $(0,0)$. To show this, we show $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist. In particular, we find $x^1_n\to 0,y^1_n\to 0$ and $x^2_n\to 0,y^2_n\to 0$, but $f(x_n^1,y_n^1)$ and $f(x_n^2,y_n^2)$ converges to different valueS.

Let $x^1_n=(\frac{1}{n})^3\to 0 ,y^1_n=(\frac{1}{n})^2\to 0$, you can check $f(x_n^1,y_n^1)\to 2$.

Let $x^2_n=(\frac{1}{n})^3\to 0 ,y^2_n=2(\frac{1}{n})^2\to 0$, you can check $f(x_n^2,y_n^2)\to 16$

Answer 2

**Hint:**As you approach $(0,0)$ along $y^3=mx^2$, you get $lim_{(x,y)→(0,0)}f(x,y)=2m/(1+m^2)$ which depends on arbitrary $m$.