Can anyone explain how to do this? Find the limit of 
$\lim\limits_{(x,\ y)\to(0,\ 0)}f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2} $, using squeeze theorem.
So far i have done is :
- |$\frac{1}{2}x$ +2|<= $\frac{x^3+4x^2+2y^2}{2x^2+y^2} $ <= |$\frac{1}{2}x$ +2|
But the limit of the LHS = -2 and RHS = 2.
The answer is 2
 A: $$f(x,y)=\frac{x}{2}+2-\frac{xy^2}{2(2x^2+y^2)}. \tag{1}$$
So we one only need show that $g(x,y)=xy^2/(2x^2+y^2) \to 0$ as $(x,y) \to (0,0).$ In the latter put $x=r \cos t, y=r \sin t,$ and get
$$r \cdot \frac{\sin^2 t \cos t}{\cos^2 t+1}.$$
Then since the fraction is bounded and $r \to 0$ as $(x,y) \to (0,0),$ the limit of $g$ is zero.
[Of course this doesn't use the squeeze theorem, at least not explicitly. Maybe it could be recast to do that.]
To reformulate using squeeze: Let $K(x,y)$ be the fraction on the right of $(1)$. Then we have $|f(x)-2|=|x/2+K(x,y)| \le |x/2|+|K(x,y)|$ sp tjat we cam "sandwich" as
$$2-|x/2|-|K(x,y)| \le f(x,y) \le 2 + |x/2| + |K(x,y)|.$$
Then each side can be shown to approach $2$ as outlined, and the squeeze theorem then arrives at the limit $2$ for $f(x,y).$
A: $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}$$
Then $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}\le \frac{x^3}{2x^2}+2 \le \frac{x}{2}+2 $$
A: There are problems with both sides of what you've done.  The correct inequalities to derive are
$$2-{1\over2}|x|\le {x^3+4x^2+2y^2\over2x^2+y^2}\le2+{1\over2}|x|$$
(after which the squeeze theorem gives the limit $2$).  A way to get this is to write
$${x^3+4x^2+2y^2\over2x^2+y^2}=2+{1\over2}x\left({2x^2\over2x^2+y^2}\right)$$
and then note that
$$0\le{2x^2\over2x^2+y^2}\le1$$
(The last expression is clearly non-negative because the variables are squared, and it's less than $1$ because the (positive) denominator is bigger than the (non-negative) numerator.)
