<s>[INST] Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the function:
$$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$
I see the denominator is always positive so this is equal to
$\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$.
Using the triangle inequality i know that this is equal or less than
$\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$.
From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$.
Im thinking i can write
$$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$
for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|.
Could someone give me any pointers?
 [/INST] Use the fact that $|xy| \le \frac{x^2 + y^2}{2}$. You can then get,
$$3|xy^5| = 3|xy|y^4 \le \frac{3}{2}(x + y)^3$$
and
$$2x^2y^4 = 2|xy|^2y^2 \le \frac{1}{2}(x + y)^3.$$
You can then proceed as you were.
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