"Sandwich" or "squeeze" rule for function 2 variables So let's say I wanted to find $$ \displaystyle \lim_{(x,y) \to  (0,0)}\dfrac{xy^2} {x^2 +y^4} $$
Let's call the function above $f(x,y)$ then  $h(x,y) = 0$ would be a function smaller than $f$ that would work for the sandwich or squeeze rule. What's an easy way to find a function larger than $f(x,y)$ in order to complete the sandwich rule? Is there some specific method? Thanks in advance.
 A: Hint:
$$
\begin{align}
\frac{xy^2}{x^2 +y^4}
&=\frac1{\frac{x}{y^2}+\frac{y^2}{x}}\\
&=\frac{\operatorname{sign}(x)}{\frac{|x|}{y^2}+\frac{y^2}{|x|}}\\
&=\frac{\operatorname{sign}(x)}{\left(\frac{\sqrt{|x|}}{y}-\frac{y}{\sqrt{|x|}}\right)^2+2}\\
\end{align}
$$
Consider the paths $(t^2,t)$ and $(-t^2,t)$ as $t\to0$.
A: The function $f(x,y)$ does not have a limit as $(x,y)\to (0,0)$. 
To see this, note that if we approach $(0,0)$ along the path $x=y$, the limit is $0$, while if we approach $(0,0)$ along the path $x=y^2$, the limit is $\frac{1}{2}$. 
In particular, if we use lower piece of bread $h(x,y)=0$, there is no $k(x,y)$ which approaches $0$ such that $h(x,y)\le f(x,y)\le k(x,y)$. 
A: One of the classical method that you can show that this kind of limits dont exist is this, you use $x=my^k$ or $y=mx^k$ where $m$ is variable and $k$ is integer. After substitution  $x=my^k$ or $y=mx^k$ in the limit, if the limit be dependent of the variable $m$, it means, it dosent limit. There are three note one) this method is acceptable in just in the case $(0,0)$. second) after substitution $x=my^k$ or $y=mx^k$, when you can say It dosent limit that in the limit dosent exist $x$ and $y$ and three) this method is necessary condition not sufficient, I mean with this method just you can show there is no limit not exist of the limit.For your question you should choose $x=my^2$ and substitute it in the limit:
$$\frac{my^2y^2}{(my^2)^2+y^4}=\frac{my^4}{m^2y^4+y^4}=\frac{m}{m^2+1}$$ 
we can see, the limit is dependent to $m$ and so there is no limit in $(0,0)$.
