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:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$

Since
$$
0\leq \frac{y^2}{x^4+y^2} \leq 1 \qquad \forall (x,y)\neq (0,0),
$$
you get
$$
|f(x,y)| = |x|^3 \left[\frac{y^2}{x^4+y^2}\right]^2 \leq |x|^3,
\qquad \forall (x,y)\neq (0,0),
$$
hence $f(x,y) \to 0$ as $(x,y)\to (0,0)$.
