Here, I have the function $u(x,y)$ = \begin{cases} \frac{x^{4/3}y^{5/3}}{x^{2} + y^{2}}, & \text{if $(x,y)\neq$ 0} \\ 0, & \text{if $(x,y)$ = 0} \end{cases}
I need to determine whether this function is continuous at $(0,0)$ and support my answer. I know how to prove it isn't continuous, by finding a limit of the first function which isn't equal to $0$, but I'm not sure how to prove that it is continuous. I feel like I should start by trying the epsilon delta proof, showing that $\lvert f(x.y) - f(0,0)\rvert$ = $\lvert \frac{x^{4/3}y^{5/3}}{x^{2} + y^{2}}\rvert \le 1$, but I'm unsure how to do this precisely.
$\endgroup$ 12 Answers
$\begingroup$If $x=r\cos t,\ y=r \sin t,$ your function becomes (when $(x,y) \neq (0,0)$ so the polar is OK) $$\frac{r (\cos t)^{4/3} (\sin t)^{5/3}}{\cos^2 t+\sin^2 t}.$$ Then as $x,y \to 0,$ the denominator is always $1$ and the absolute value of the numerator bounded above by $|r|.$
$\endgroup$ $\begingroup$The other method is to write $y=mx$.
Note : $m=m(x,y)$ is variable, it is the ratio of converging rates of $y$ and $x$.
$\displaystyle{f(x,y)=\frac{m^{\frac 53}x^3}{x^2(1+m^2)}=\frac{m^{\frac 53}x}{1+m^2}}$
If $m$ is bounded then $|m|<C$ and $|\frac{1}{1+m^2}|<1$ thus $|f(x,y)|\le C^{\frac 53}|x|\to 0$
If $|m|\to+\infty$ then $f(x,y)\sim\frac{x}{m^{\frac13}}\to 0$
So in all cases $f(x,y)\to 0$ when $(x,y)\to (0,0)$
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