$\lim\limits_{n\to\infty} \dfrac{0}{\left(\frac{2}{n^2}\right)^2}\overset{?}{=}0$. $n=\infty$ is not a case I need to consider, right?

Question

Consider the sequence defined by $(x_n, y_n)=(\frac{1}{n}, \frac{1}{n}),\quad n\in \mathbb{N}$ and the function $f(x,y)=-\dfrac{y^4+x^3y-xy^3-x^4}{\left(y^2+x^2\right)^2}$ Now I want to calculate the limit of $$\lim\limits_{n\to\infty}f((x_n, y_n))=-\lim\limits_{n\to\infty} \dfrac{\frac{1}{n^4}+\frac{1}{n^4}-\frac{1}{n^4}-\frac{1}{n^4}}{\left(\frac{2}{n^2}\right)^2}=-\lim\limits_{n\to\infty} \dfrac{0}{\left(\frac{2}{n^2}\right)^2}\overset{?}{=}0$$ I'm unsure about the last step, because $\lim\limits_{n\to\infty} \left(\frac{2}{n^2}\right)^2 =0$. I think $\lim\limits_{n\to\infty} \dfrac{0}{\left(\frac{2}{n^2}\right)^2}=\frac{0}{0}$ would be wrong. I can treat $\dfrac{0}{\left(\frac{2}{n^2}\right)^2}$ like the constant zero-sequence? Because for every $n$:$\dfrac{0}{\left(\frac{2}{n^2}\right)^2}=0$ and $n=\infty$ is never "reached", correct? I hope you understand my question, I had difficulty formulating it.

Answer 1

Numerator is excatly $0$

Denominator is approaching $0$

So $\frac{0}{\text{value near 0 but not 0}} = 0$

Answer 2

Used an alternative route got $0$ as well.

$\lim _{(x,y)\to (0,0)} - \frac{y^4+x^3y-xy^3-x^4}{(x^2+y^2)^2}=L$

Let $y=mx$

$L= -\frac{m^4+m-m^3-1}{(1+m^2)^2}=-\frac{(m^3+1)(m-1)}{(1+m^2)^2}$

$\frac{dL}{dm}\ne 0$, so the limit does not exist, it is determined by which path you use to get to the origin.

$m=1\implies L=0$ is consistent with your scenario.

I think you'll get the same result even off the origin. Let $x=a +\frac{1}{n}, y = b + \frac{1}{n}$

$m=b/a\implies L= - \frac{(b^3/a^3+1)(b/a-1)}{(1+b^2/a^2)^2}= - \frac{(b^3+a^3)(b-a)}{(a^2+b^2)^2}$.