Find $\lim_{(x,y)\to(0,0)}\frac{(x^2-y)}{(4x^2+y^2)}$ using ε–δ definition of a limit I was wondering if anyone could help with this $\epsilon–\delta$ definition of a limit. I have looked it up in my calculus book and online and I just don't understand how to do it.
Prove, using the  $\epsilon–\delta$ definition of a limit that
$$\lim_{(x,y)\to(0,0)}\frac{(x^2-y)}{(4x^2+y^2)}$$
 A: The limit doesn't exist since $$\lim_{t\to 0}f(t,0)=\frac{1}{4}$$ and $$\lim_{t\to 0^+}f(0,t)=-\infty .$$
A: The ϵ–δ definition of a limit is not used to find a limit, but to prove that a certain (previously known) candidate is a limit indeed (or to prove that is not, given the case). 
So what you have to do first is to have a candidate for your limit, and then use the ϵ–δ definition to prove that it indeed is.
So first things first: find the candidate for your limit so that then you can use the ϵ–δ definition. The usually most straightforward method is applying what is called 'direct substitution' (i.e. plugging in the values under the 'lim' word in the formula on the right of 'lim'). If that results in a number, you proceed with the ϵ–δ definition. But sometimes this doesn't work, like in this case, because you get $ \frac{0}{0} $.
When this happens, you have to 'evaluate trajectories' to get your candidate. However, take notice that in the process of doing this, you might actually find that for certain limits, different trajectories result in different candidates. Which of course means that the limit in that case doesn't exist, because it must only be one regardlessly of the trajectory chosen.
This is what happens in your case as shown by user @Surb.
So, the limit doesn't exist in your case for sure. Therefore you can't use the ϵ–δ to prove that any candidate is a limit. You can still use the ϵ–δ to prove that the limit doesn´t exist, if you want to. Here is explained how to do it.
