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I'm trying to show that $$\lim_{(x,y)\to (0,0)} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0$$
I've used polar coordinates but when I do this I get the possibility of $\frac{0}{0}$ if $\cos(\theta)\to 0$ as $r\to 0$. So it must be that I need some sort of bound to make sense of this limit. But I'm not sure how to proceed.
Hint: $u^2 + 4v^2 = (u-2v)^2 + 4uv \ge 4uv$, so:
$$x^6 + 4y^8 = (x^3)^2 + 4(y^4)^2 \ge 4x^3 y^4$$
Edit:
I pointed out in a comment that an argument by symmetry works.
A slightly better approach might be as follows:
$$\left| \frac{12x^3 y^5 + 4 x^4y^4}{x^6 + 4y^8} \right| \le 4\frac{3|x|^3 |y|^5 + x^4y^4}{x^6 + 4y^8}$$
Now we use $x^6 + 4y^8 \ge 4|x|^3 y^4$ to see that the RHS $\to 0$.
If we wish to use polar coordinates $(r,\theta)$, then we can write
$$\begin{align} \frac{12x^3y^5+4x^4y^4}{x^6+4y^8}&=r^2\left(\frac{12\cos^3(\theta)\sin^5(\theta)+4\cos^4(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\\\\ &=r\left(3\sin(\theta)+\cos(\theta)\right)\left(\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}\right)\tag1 \end{align}$$
Let $ g(r,\theta)=\frac{4r\cos^3(\theta)\sin^4(\theta)}{\cos^6(\theta)+4r^2\sin^8(\theta)}$. Denote $\sin(\theta)$ by $s$ and $\cos(\theta)$ by $c$.
We will view $g(r,\theta)$ as a function of $\theta\in \mathbb{R}$, which is differentiable and $2\pi$-periodic. Therefore the extrema occur at points for which $\frac{\partial g(r,\theta)}{\partial \theta}=0$.
Then, taking the partial derivative with respect to $\theta$, and , we have
$$\begin{align} \frac{\partial g(r,\theta)}{\partial \theta}&=4rs^3c^2\,\left(\frac{(4c^2-3s^2)(c^6+4r^2s^8)-s^2c^2(32r^2s^6-6c^4)}{(c^6+4r^2s^8)^2}\right)\\\\ &= 4rs^3c^2\,\left(\frac{(3+c^2)(c^6-4r^2s^8)}{(c^6+4r^2s^8)^2}\right)\tag 2 \end{align}$$
We see that $\frac{\partial g(r,\theta)}{\partial \theta}=0$ when $\sin(\theta)=0$ or $\cos(\theta)=0$ or $\cos^6(\theta)=4r^2\sin^8(\theta)$.
When $\sin(\theta)=0$ or $\cos(\theta)=0$, $g(r,\theta)=0$. When $\cos(\theta)^6=4r^2\sin^8(\theta)$,
$$g(r,\theta)=\text{sgn}(\cos(\theta))\tag 3$$
Finally, using $(3)$ in $(1)$ reveals
$$\left|\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}\right|\le r|3\sin(\theta)+\cos(\theta)|$$
whereupon applying the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{12x^3y^5+4x^4y^4}{x^6+4y^8}=0}$$
