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

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.)
