I'm trying to determine which values of $\alpha$ and $\beta$ make the following function continuous:
$$f(x,y)=\left\{
\begin{array}{c}
\frac{|{x}|^{\alpha}|{y}|^{\beta}}{x^4 + y^4} \quad \text{ if }(x,y) \neq (0,0) \\
0 \quad \text{ if } (x, y)=(0,0)
\end{array}
\right.$$
$f$ is clearly continuous in $\mathbb{R}^2 – \{(0,0)\}$, and for $f$ to be continuous in $(0,0)$ the limit
$$\lim_{(x,y)\to(0,0)}\frac{|x|^{\alpha}|y|^{\beta}}{x^4 + y^4}$$
has to exist and be equal to zero.
Points where $f$ is not continuous:
Using the sequencial characterization of limits, if we consider $\{x_k\}=\{\frac{1}{k}, \frac{1}{k}\}$, it converges to $(0,0)$ and $x_k \neq (0,0) \ \forall k \in \mathbb{N}$, so if $f$ is continuous in $(0,0)$ then $\{f(x_k)\}$ must converge to zero.
$$f(\frac{1}{k},\frac{1}{k})=\frac{|\frac{1}{k}|^{\alpha}|\frac{1}{k}|^{\beta}}{(\frac{1}{k})^4 + (\frac{1}{k})^4}=\frac{k^4}{2{k}^{\alpha + \beta}}$$
If $\alpha+\beta \leq 4$ then $f(\frac{1}{k},\frac{1}{k})$ does not converge to zero and therefore $f$ is not continuous.
I don't know how to handle the cases where $\alpha + \beta \gt 4$ and either $\alpha$ or $\beta$ are less than two. Any ideas? Or maybe a quicker way to approach this problem? Thanks.
Best Answer
Let:
$$A=\{(x,y) \in \mathbb{R}^2: \text {max} \{|x|,|y|\}=|x| \}$$ $$B=\{(x,y) \in \mathbb{R}^2: \text {max} \{|x|,|y|\}=|y| \}$$
Note that $A \cup B = \mathbb{R}^2$.
If $(x,y) \in A$:
$$0 \leq \frac{|x|^{\alpha}|y|^{\beta}}{x^4 + y^4} \leq \frac{|x|^{\alpha+\beta}}{x^4 + y^4} \leq \frac{|x|^{\alpha+\beta}}{x^4} = |x|^{\alpha+\beta-4}$$
So if $\alpha + \beta \gt 4$ then:
$$0 \leq \lim_{(x,y)\to(0,0)}\frac{|x|^{\alpha}|y|^{\beta}}{x^4 + y^4} \leq \lim_{x\to0} |x|^{\alpha+\beta-4} = 0$$
After using the same procedure in $B$ we conclude that: