$f(x,y)=\frac{|x|^k |y|^k}{x^2+y^2}$ when $(x,y)\neq (0,0)$ and $f(0,0)=0$. Find the smallest $k$ such that $f$ is continuous at $(0,0)$.

Question

$f(x,y)=\frac{|x|^k |y|^k}{x^2+y^2}$ when $(x,y)\neq (0,0)$ and $f(0,0)=0$. Find the smallest $k$ such that $f$ is continuous at $(0,0)$.

Approaching $(0,0)$ from $x-axis$ or $y-axis$, we always get the limit to be zero.

Approaching (0,0) from $y=x$, then $f(x,x)=\frac{|x|^{2k}}{2x^2}= \frac{x^{2k}}{2x^2}=\frac{1}{2}x^{2k-2}$

If $2k-2> 0$, then $\frac{1}{2}x^{2k-2}\to 0$ when $x\to 0$. Hence, $k$ should at least be greater than 1?

There is no smallest $k$. The function is continuous at $(0,0)$ if and only if $k>1$. — Did, Mar 22 '17 at 13:52
@Did How to see $k>1$? I only choose to approach (0,0) from $y=x$. What about other direction? — Kenneth.K, Mar 22 '17 at 13:53
Use $2|x|,|y|\leqslant x^2+y^2$ and conclude that $f(x,y)\leqslant\frac12|x|^{k-1}|y|^{k-1}$. — Did, Mar 22 '17 at 13:56
If I am closing in on f by trajecting the curve $y = x^{\frac{1}{k}}$, don't I get $\frac{0}{0}$ at $(0,0)$ — meguli, Mar 22 '17 at 19:59
Possible duplicate of Multivariable limit proof: $\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$ — Arnaud D., Dec 04 '19 at 13:42

score 2 · Accepted Answer · answered Mar 22 '17 at 13:55

2

Yes, $k$ should at least be greater than $1$. With polar coordinates:

$f(r \cos t, r \sin t)=\frac{r^{2k}|\cos^kt \sin^k t|}{r^2}=r^{2k-2}|\cos^kt \sin^k t|$.

Its now your turn to show:

$ \lim_{(x,y) \to (0,0)}f(x,y)=0 \iff k>1$

answered Mar 22 '17 at 13:55

Fred

1 Answers1