Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist

Question

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:

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

Answer 1

Let $u=x^4+y^2$. Then $|x|\le u^{1/4}$ and $|y|\le u^{1/2}$. So $|x^3y^4|\le u^{11/4}$ and, for $(x,y)\ne(0,0)$, $|f(x,y)|\le u^{3/4}$.

Answer 2

Using that for $y \neq 0$:

$$|\frac{x^3y^4}{(x^4+y^2)^2}| \leq |\frac{x^3y^4}{(0+y^2)^2}|=|x^3| \to 0$$

As a smaller denominator means larger number in magnitude. We easily conclude by squeeze theorem that the limit is $0$.

Answer 3

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.

$\displaystyle f(x,y)=\frac{x^3u^4x^8}{(x^4+u^2x^4)^2}=\underbrace{\frac{u^4}{(1+u^2)^2}}_\text{bounded}\ \underbrace{x^3}_{\to 0}\to 0$

Answer 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 $(x,y)\to (0,0)$.