<s>[INST] Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$ 
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
  exists. If it does, then calculate its value.

My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$
But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$
Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$
So the left hand and right hand limits cannot be equal, and therefore the limit does not exist. 
 [/INST] Consider $$f(x,y)=\frac{(x+y)^2}{x^2+y^2} .$$
If you take the path $(0,y)$, then: 
$$\displaystyle\lim_{y\to0} f(0,y) =\lim_{y\to0} \frac{y^2}{y^2}=1$$
If you take the path $(x,x)$, then: 
$$ \lim_{x\to0}f(x,x) =\lim_{x\to0} \frac{(2x)^2}{2x^2}=2$$
So, the limit doesn't exist.
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