# Find multivariable limit.

I need a hint on what kind of path I should choose to show that this limit does not exist:

$\displaystyle\lim_{(x,y) \to (0,2)}\frac{(xy - 2x)}{x^3 + (y - 2)^2}$

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Did you try $(0,y)$ and $(x, x^2+2)$? – Sigur Dec 8 '12 at 20:09
You can simplify things a bit by noting your expression is the same as $\lim\limits_{(x,y)\rightarrow(0,0)}{xy\over x^3+y^2}$. – David Mitra Dec 8 '12 at 20:13
I would of never had thought of putting x^2 + 2 into y. What was your thought process? Are there some patterns or techniques? – cRaZiRiCaN Dec 8 '12 at 20:14
Try to evaluate the limit along the path $$y = 2+x^{3/2}$$ and then evaluate the limit first letting $x \to 0$ and then $y \to 2$. – user17762 Dec 8 '12 at 20:14
what about straight lines?$y-2=kx$? the limit on these should be $1/k$... – wisefool Dec 8 '12 at 20:31

We have $$\begin{split} \displaystyle\lim_{(x,y) \to (0,2)}\frac{(xy - 2x)}{x^3 + (y - 2)^2} & = \displaystyle\lim_{(x,y) \to (0,2)}\frac{x(y - 2)}{x^3 + (y - 2)^2} \end{split}$$
If $f(x,y)=\frac{xy)}{x^3 + y^2}$ and $\varphi(x,y)=(x,y-2)$ then $$\displaystyle\lim_{(x,y)\to(0,+2)}\varphi(x,y)=(0,0)$$ and $$\displaystyle\lim_{(x,y)\to(0,0)}f(\varphi(x,y))= \displaystyle\lim_{(x,y) \to (0,2)}\frac{x(y - 2)}{x^3 + (y - 2)^2}$$ But $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ don't exist, by cause if $x=y=t$ we have $$\displaystyle\lim_{(x,y) \to (0,0)}\frac{xy}{x^3 + y^2} = \displaystyle\lim_{t \to 0}\frac{t^2}{t^3 + t^2} = \displaystyle\lim_{t \to 0}\frac{1}{t + 1}=1$$ and if $x=t$, $y=2t$ we have $$\displaystyle\lim_{(x,y) \to (0,0)}\frac{xy}{x^3 + y^2} = \displaystyle\lim_{t \to 0}\frac{2t^2}{t^3 + 4t^2} = \displaystyle\lim_{t \to 0}\frac{2}{t + 4}=\frac{1}{2}$$