I'm working on this problem:
Let: $$ f(x,y) = \begin{cases} \frac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}} &\text{if} \ (x,y) \neq (0,0)\\ 0 &\text{if} \ (x,y) = (0,0) \end{cases} $$
I want to show that $f$ is continuous at the origin.
So my scratch/set up is:
$\forall\varepsilon>0, \exists\delta>0 \ s.t. \ |(x,y)-(\overline{x},\overline{y})|<\delta \implies |f(x,y) - f(\overline{x},\overline{y})|<\varepsilon$
So I get:
$|f(x,y) - f(\overline{x},\overline{y})|=\left|{\cfrac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}}-0}\right|<\varepsilon$
If we convert to polar coordinates and utilise the Pythagorean Identity:
$x = r\cos{\theta}, \ \ \ y = r\sin{\theta}$
The function should reduce down to (an image of the steps taken are below):
$\left|{\cfrac{x^{3/5}y^{2/5}}{\sqrt{x^2+y^2}}-0}\right|=(\cos^{3}{\theta}\cdot\sin^{2}{\theta})^{1/5}<\varepsilon$
If we were working in the (x,y) plane the sinusoidal functions would be bounded above by 1, however I'm not 100% sure what they are bounded by in polar form. Am I on the right track for showing continutiy?
SCRATCH WORKINGS:
You arrive at the conclusion that for any $(x, y) = (r\cos\theta, r\sin\theta)\neq (0, 0)$, you have that $$ f(x, y) = (\cos^3\theta\cdot \sin^2\theta)^{1/5} $$ for arbitrary $\theta$. What happens if I choose $\theta = \pi/4$? What about $\theta = 0$? What does this tell you about the continuity at the point $(0, 0)$?