How to determine the limit of $f(\mathbf{x}) = \frac{2x^6y}{x^8+y^4+5x^4y^2}$ as $\mathbf{x} \rightarrow \mathbf{0}$? How do I determine the limit of $f(\mathbf{x}) = \frac{2x^6y}{x^8+y^4+5x^4y^2}$ as $\mathbf{x} \rightarrow \mathbf{0}$, where $f:\mathbb{R^2}-\{(0,0)\}\rightarrow \mathbb{R}$?
Following the suggestion I received in this answer, I tried to convert the function into polar coordinates, and this is what I got:
$$ f(r,\theta) = \frac{r^32\cos^6(\theta)\sin(\theta)}{r^4\cos^8(\theta)+\sin^4(\theta)+5r^2\cos^4(\theta)\sin^2(\theta)}$$
I am not aware of how the above expression can be further simplified further. What should I do next?
Later on the question also asks if $f(\mathbf{x})y$ has a limit as $\mathbf{x}\rightarrow 0$.
Please advise on the method that I should use to solve this kind of problem.
EDIT:
The solution given is:


*

*$f(\mathbf{x})$ has no limit as $\mathbf{x}\rightarrow \mathbf{0}$

*the limit of $f(\mathbf{x})y$ is $0$ as  $\mathbf{x}\rightarrow \mathbf{0}$


The solution simply lists the steps to show why the above is true. It does not discuss the approach. In fact, it starts by saying some experimentation is usually required and did not elaborate further. As such, if I was faced with a similar question in the future, I doubt that I will be able to make progress.
Therefore, how should I approach this sort of question i.e. to determine the limit of a two variable rational function as it tends to some point. What observations should I make to determine the right course of action? Even if there is no set way, what heuristics can I follow to help arrive at the answer?
I've not seen any text discuss a strategy for these kind of questions, so I would also appreciate it if someone can direct me to an online resource (if available). If it helps to narrow the scope of the answer, I am doing an introductory real analysis course.
 A: Because in the denominator there's a symmetry between $y$ and $x^2$, it's natural to look at the behavior of the function along parabolas $y = ax^2$. Then your function becomes
$$f(x,ax^2) = {2ax^6x^2 \over x^8 + (ax^2)^4 + 5x^4(ax^2)^2}$$
$$= {2a \over 1 + a^4 + 5a^2}$$
So $f(x,y)$ is constant on each such parabola, but the constant varies with $a$. Since all of these parabolas contain the origin, you get different limits as you approach the origin along each parabola. Hence the overall limit doesn't exist.
A: This expression does not have a limit as ${x}$ tends to zero, and hence no method you try to use will not work.
In order to show it has no limit you only have to come up with two different paths by which you can reach two different limits.
Try looking at $y=0$ identically and $x \rightarrow 0$ on one hand.
And at $y=x^2$, $x\rightarrow 0$ on the other hand.
A: Since $y$ cannot be zero then
$$\lim_{x\rightarrow0}\frac{2x^6y}{x^8+y^4+5x^4y^2}=\frac{0}{y^4}=0$$
Multiplying $f$ by $y$ makes no difference to the limit.
