<s>[INST] Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$ if$\ δ=ε$ we have to prove that:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|<ε$$
But I'm having a hard time trying to prove the last part, I tried:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$
 [/INST] Let's rewrite $y=ux^2$ in order to render the denominator homogeneous.
Note: $u$ is variable, write it $u=\frac y{x^2}$ if you prefer.
$\displaystyle f(x,y)=\frac{x^3u^4x^8}{(x^4+u^2x^4)^2}=\underbrace{\frac{u^4}{(1+u^2)^2}}_\text{bounded}\ \underbrace{x^3}_{\to 0}\to 0$
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