Find $\lim_{(x,y)\in S} \frac{x^2+ay^2}{x^2+by^2}$ as $(x,y) \to (0,0)$ for specified $S$. Find $$\lim_{\stackrel{(x,y)\to(0,0)}{(x,y)\in S}} \frac{x^2+ay^2}{x^2+by^2}$$ where $a,b\in \mathbb{R}$ and


*

*$S=S_1=\{(x,y)\in\mathbb{R}^2 \mid |y| < c|x|^p\}$ with $c > 0, p > 1$

*$S=S_2=\{(x,y)\in\mathbb{R}^2 \mid |y| < |x|\}$.
Here's my attempt by changing to polar coordinate.
$$x = r\cos \theta,\quad y = r\sin \theta$$.
So $$
\begin{align*}
f(x,y) = \frac{x^2+ay^2}{x^2+by^2} &= \frac{r^2(\cos^2\theta+a\sin^2\theta)}{r^2(\cos^2\theta +b\sin^2\theta)} \\
&=\frac{1+a + (1-a)\cos 2\theta}{1+b+(1-b)\cos 2\theta}.
\end{align*}
$$
For $S_1$, I think $|y| < c|x|^p$ "forces" $\theta \to 0$ as $(x,y)\to 0$. So $\cos 2\theta \to 1$ and $f(x,y) \to 1$.
For $S_2$, $\theta$ could be anything in $[-\pi, -\pi+\pi/4) \cup (-\pi/4,\pi/4) \cup (\pi -\pi/4, \pi)$. So the limit does not exist.
How can I make these rigorous?
Thank you!

Set $$f(x,y)=\frac{x^2+ay^2}{x^2+by^2}$$  I don't think polar coordinates help.
For starters, the limit of $f(x,0)$ as $x\to 0$ is then $1$, so if the limit of $f$ exists for $(x,y)\to (0,0)$, it must be $1$.
For case 1, $|y|<c|x|^p=o(x)$ (i.e., $|y|$ goes to zero faster than $x$) in $S_1$. We have also $ay^2=o(x^2)$ and $by^2=o(x^2)$ as $(x,y)\to (0,0)$, so the limit of $f$ as $(x,y)\to(0,0)$ is $$\lim_{x\to 0} {x^2+o(x^2)\over x^2+o(x^2)} = 1$$ as desired.
For case 2, let's suppose $y=x/c$, $c>1$, so that the line $y=x/c$ lies within $S_2$.  We then have $x^2+ay^2=x^2(1+a/c^2)$ and likewise $x^2+by^2=x^2(1+b/c^2)$ so the value of $f$ on the line $y=x/c$ is $(1+a/c^2)/(1+b/c^2)$.  Since $c>1$ is arbitrary, the value of $f$ on different lines in $S_2$ passing through the origin is different, and so the limit of $f$ as $(x,y)\to (0,0)$ doesn't exist.
