Is this a correct use of the squeeze theorem? I have to find the following limit:
$$
\lim_{x\to0,y\to0} \frac{x^2y^2}{x^2+y^4}=[\frac{0}{0}]
$$
I try to reach the origin moving on the y-axis ($x=0$):
$$
\lim_{y\to0} \frac{0}{y^4}=0
$$
I get the same result when I am moving on the x-axis ($y=0$):
$$
\lim_{x\to0} \frac{0}{x^2}=0
$$
The same if I move over ($y=x$):
$$
\lim_{y\to0} \frac{x^2x^2}{x^2+x^4}=\lim_{y\to0} \frac{x^2}{1+x^2} = 0
$$
So I start to get confident that the limit should be 0. Let's prove it.
$$
\left| f(x,y) - l \right| = \left| f(x,y) - 0 \right| =  \left| \frac{x^2y^2}{x^2+y^4} \right| 
$$
Looking at the fraction we know that:
$$
x^2 \leq x^2+y^4
$$
So...
$$
\frac {x^2} {x^2+y^4} \leq 1 \forall (x,y) \neq (0,0)
$$
We can now say that:
$$
\left|y^2\frac{x^2}{x^2+y^4}\right|\leq \left|y^2\right|
$$
So for the squeeze theorem we have to get:
$$
-h(x,y) \leq f(x,y) \leq h(x,y)
$$
And..
$$
\lim_{x\to0,y\to0} h(x,y) = \lim_{x\to0,y\to0} y^2 = 0
$$
So we can say that the limit is 0. Is this prove right? Are there other kind of prove I can learn?
 A: Looks fine to me.
Another general way to do that (even though in this case your way is probably the best one )is to set $x = \rho \cos \theta$, $y = \rho \sin \theta$ and taking the limit as $\rho \to 0$.
There are some catches with this techniques though, as you should first find a function of $\rho$ which is in modulus bigger than the other one and show that it goes to $0$.
For example in this case, setting $f(x, y) = \frac{x^2y^2}{x^2+y^4}$, you try to find a function $g(\rho)$ such that $$|f(\rho \cos \theta, \rho \sin \theta)| \le g(\rho) \to 0$$ and thanks to the squeeze theorem you can conclude. The calculations in your case are  (notice that I did not put the modulus as all quantities are positive)
$$ \frac{\rho ^ 4 \sin^2 \theta \cos ^2\theta}{\rho^2 (\cos ^2 \theta + \rho^2 \sin^4 \theta)} = \frac{\rho ^ 2 \sin^2 \theta \cos ^2\theta}{\cos ^2 \theta + \rho^2 \sin^4 \theta} \le \frac{\rho ^ 2 \sin^2 \theta \cos ^2\theta}{\cos ^2 \theta} = \rho^2 \sin ^2 \theta \le \rho^2 = g(\rho) \to 0 $$
And, again thanks to the squeeze theorem, you can conclude that the limit is $0$
A: The proof is correct. The squeeze theorem is one of the most used technique to prove 2D limits of functions. 
