<s>[INST] 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
 [/INST] $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}$$
Then $$f(x,\ y) = \frac{x^3+4x^2+2y^2}{2x^2+y^2}=\frac{x^3}{2x^2+y^2}+\frac{2(2x^2+y^2)}{2x^2+y^2}\le \frac{x^3}{2x^2}+2 \le \frac{x}{2}+2 $$
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