<s>[INST] I need a sanity check to compute the limit of a function in two variables So I have $g(x,y) = \tfrac{f(x,y) - y}{\sqrt{x^2 + y^2}}$, where $f(x,y) = \tfrac{y^3 -x^8y}{x^6 + y^2}$ for $(x,y) \neq (0,0)$.
I want to find $\lim_{(x,y) \to (0,0)} g(x,y)$ if it exists.
My question is if after doing some algebraic manipulation, I can get g as a product of $\tfrac{-y}{\sqrt{x^2 + y^2}}$ and $\tfrac{x^6(1+x^2)}{x^6 + y^2}$, can I bound the former by $1$ in absolute value? What can I do with the second part of the product? I feel that the numerator would dominate but want to be rigorous with it. Any help would be appreciated.
 [/INST] call the ratio $G,$  I got
$$ |G| = \frac{|y| x^6 (1+x^2)}{r (x^6 + y^2)}  $$
Demand $c = | \cos \theta | $  and $s = | \sin \theta | $
$$ |G| = \frac{r^7 c^6 s (1+r^2 c^2 )}{r^3 (r^4c^6 +s^2)}  $$
$$ |G| = \frac{r^4 c^6 s (1+r^2 c^2 )}{ r^4c^6 +s^2}  $$
We are allowed to demand $r < 1,$ so that $1+r^2 c^2 < 2,$
$$ |G| < 2 \frac{r^4 c^6 s }{ r^4c^6 +s^2}  $$
From $$ 0 \leq (r^2 c^3 \pm s)^2 $$
we get $$ r^4 c^6 + s^2 \geq 2 r^2 c^3 s   $$  and
$$ \frac{1}{r^4 c^6 + s^2} \leq \frac{1}{2 r^2 c^3 s}   $$
$$ |G| < 2 \frac{r^4 c^6 s }{ r^4c^6 +s^2} \leq 2 \frac{r^4 c^6 s }{2 r^2 c^3 s } \leq r^2 c^3 \leq r^2$$
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