I've been struggling with this exercise. Edited For every $\alpha, \beta, \gamma, \delta \gt0 \in \mathbb R$, compute, if it exists, $$ \lim_{(x, y)\to (0,0)} \frac {{\vert x \vert}^\alpha+ {\vert y \vert}^\beta}{{\vert x\vert}^\gamma + {\vert y \vert}^\delta} $$ In my opinion, if $\alpha=\beta$ and $\gamma=\delta$, the limit exists and can be 0, 1 or $\infty$ depending on $\alpha-\gamma$. More specifically, if $\alpha=\beta$ and $\gamma=\delta$ we have $$\alpha\gt \gamma \Rightarrow\lim_{(x, y)\to (0,0)}{f(x,y)}=0 $$ $$\alpha= \gamma \Rightarrow\lim_{(x, y)\to (0,0)}{f(x,y)}=1 $$ $$\alpha\lt \gamma \Rightarrow\lim_{(x, y)\to (0,0)}{f(x,y)}=\infty $$.
Then we can say that if $\alpha \gt \gamma $ and $\beta \gt \delta $, using polar coordinates in the plane, triangle inequalities and company, $${\vert f(x,y) \vert} \le {\vert \frac{{\rho^\alpha}{\vert \cos(t) \vert}^\alpha}{{\rho^\gamma}{\vert \cos(t) \vert}^\gamma} \vert}+ {\vert \frac{{\rho^\beta}{\vert \sin(t) \vert}^\beta}{{\rho^\delta}{\vert \sin(t) \vert}^\delta} \vert}\le \rho^{\alpha-\gamma}{{\vert \cos(t) \vert}^{\alpha-\gamma}}+ \rho^{\beta-\delta}{{\vert \sin(t) \vert}^{\beta-\delta}}$$, thus the limit is 0 by the squeeze theorem.
Else, if $\alpha \lt \gamma $ and $\beta \lt \delta $ we can do the same as before (changing the roles of $\alpha, \gamma$ and $\beta, \delta$), applying the squeeze theorem to $\vert \frac {1}{f(\rho, t)} \vert \le 0 $ and prove lim=$\infty$.
Moreover, if $ (\alpha-\gamma)(\beta-\delta)\lt 0 $, our limit does not exist, because, restricting our function to the axis X and Y, we have $\lim_{x\to 0} {\vert x\vert}^{\alpha - \gamma} \neq \lim_{y\to 0} {\vert y\vert}^{\beta - \delta}$ . Are there any flaws or mistakes? Is there something I'm not considering? Many thanks