Let $S$ be the set of $(\alpha, \beta) \in \mathbb{R}^2$ such that $\frac{x^{\alpha}y^{\beta}}{\sqrt{x^2+y^2}} \to 0$ as $(x, y) \to 0$. Then $S$ is contained in
$\{(\alpha, \beta): \alpha , \beta > 0\}$.
$\{(\alpha, \beta): \alpha , \beta > 2\}$
$\{(\alpha, \beta): \alpha + \beta > 1\}$.
$\{(\alpha, \beta): \alpha + 4\beta > 1\}$
As- 3
Put $y = mx$, we get $$\frac{x^{\alpha}y^{\beta}}{\sqrt{x^2+y^2}} = \frac{x^{\alpha + \beta -1}m^{\beta}}{\sqrt{1+m^2}} \to 0 $$ provided $\alpha + \beta > 1$ whenever $(x, y) \to 0$. Thus, $$S=\{(\alpha, \beta)\in \mathbb{R}^2, \ \alpha + \beta > 1\}$$
Incorrect as $S$ contains negative $\alpha$ and $\beta$'s.
Incorrect by $(1)$.
Correct as for any $\alpha, \beta)\in \mathbb{R}^2$ with $\alpha + \beta > 2$ implies $\alpha + \beta > 1$.
Incorrect as $(3, -1)$ is in the set $S$ but not in the set $\{(\alpha, \beta): \alpha + 4\beta > 1\}$.
Is my solution correct or not?
Using that $$\sqrt{x^2+y^2}\geq \sqrt{2}|x|^{1/2}|y|^{1/2}$$ we get $$\frac{|x|^{\alpha}|y|^{\beta}}{\sqrt{2}|x|^{1/2}|y|^{1/2}}$$ we get
$$\le \frac{1}{\sqrt{2}}|x|^{\alpha-1/2}|y|^{\beta-1/2}$$ so we obtain $$\alpha+\beta>1$$