Another question on the procedure of majorization I know I already asked a question about majorizations, but trust me when I say I really need to get this. Please don't whip me too much.
Let's say I need to prove that the function $$ f(x, y) = \begin{cases} \frac{x^6+y^4}{x^2+y^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0)\end{cases}$$
is continuous at the origin.
I have to stick to the distance method, hence a majorzation that leads me to conclude that $$|f(x, y) - f(0, 0)| \leq \phi(d)$$
where $\phi(d) \to 0$ as $d\to 0$, a function of a distance (or norm).
Here again is where I struggle in understanding how to operate. I mean look at this: I can either do:
$$\bigg|\frac{x^6+y^4}{x^2+y^2}\bigg| = \frac{x^6+y^4}{x^2+y^2} \leq \frac{x^2+y^2}{x^2+y^2} = 1$$
(since I am near the origin, then $x^6 < x^2$ and so on), or I can have this
$$\bigg|\frac{x^6+y^4}{x^2+y^2}\bigg| = \frac{x^6+y^4}{x^2+y^2} \leq \frac{x^3+y^3}{x^2+y^2} \to 0$$
Or also other things like
$$\bigg|\frac{x^6+y^4}{x^2+y^2}\bigg| = \frac{x^6+y^4}{x^2+y^2} \leq \frac{1}{x^2+y^2} \to +\infty$$
I still struggle in understanding how to operate. I know that it's like a very stupid question, shame on me. I just need a little push.
 A: Once you have any bound of the form $f(x,y) \le g(x,y)$, and you know $g(x,y)\to 0$, you are done, because all you want to show is continuity at the origin.
All the estimates you gave, in the forms of upper bounds are true, but some are stronger than others. For example, you identified that as $x,y\to 0$, $x^6\le x^3$ and $y^4 \le y^3$, so one bound you can show is $f(x,y)\le (x^3 + y^3)/(x^2 + y^2)$. This bound is enough to show $f(x,y)\to 0$. And it's a stronger estimate than saying $f(x,y) \le 1$, which isn't sufficient for your purpose.
To summarize, there isn't a single correct estimate, but rather there are sufficient and insufficient estimates. $f(x,y)\le (x^3+y^3)/(x^2+y^2)$ is sufficient, while $f(x,y)\le 1/(x^2+y^2)$ and $f(x,y)\le 1$ are insufficient.

To be sure an estimate is sufficient for your purpose, you will have to give a proof that it is sufficient. For example, let's work with the cubic estimate. In the typical $\epsilon,\delta$-framework, we are given $\epsilon$, and need to provide a $\delta$ so that if $|(x,y)|<\delta$, $f(x,y)<\epsilon$. The aim is to use our cubic estimate to facilitate this. So here it goes.
Suppose $|(x,y)|<\delta$, for some $\delta$ to be determined later. Then $x<\delta$ and $y < \delta$, so we have
$$
f(x,y) \le \frac{x^3 + y^3}{x^2 + y^2} \le  \delta\frac{x^2 + y^2}{x^2 + y^2} = \delta.
$$
We used our cubic estimate in the first inequality, and the assumption that $|(x,y)|\le \delta$ in the second inequality. Now it's evident that if we take $\delta = \epsilon$, $f(x,y) < \epsilon$  provided $|(x,y)|<\delta$. By the definition of limit in terms of $\epsilon,\delta$, $f(x,y)\to 0$ as $(x,y)\to 0$.
This is the only justification that our estimate "works"!
