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Find the limit of $$\exp\left(\frac{|x-2y|}{(x-2y)^2}\right)$$ when $(x,y) \to (2y,y)$.

I have considered two cases: $(x-2y)<0 $ and $(x-2y)>0$. But in first case the limit turns out to be $0$ and in the second case limit is undefined. I am not sure if my solution is correct or not.

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    $\begingroup$ hint $(x-2y)^2=|x-2y|^2$ $\endgroup$ – dmtri Dec 15 '18 at 8:36
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    $\begingroup$ $e^{\frac{1}{0}}=\infty$ if we go frrom the right of $0$ $\endgroup$ – dmtri Dec 15 '18 at 8:39
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It is :

$$\lim_{(x,y) \to (2y,y)} \exp\left({\frac{|x-2y|}{(x-2y)^2}}\right) = \lim_{(x,y) \to (2y,y)} \exp\left({\frac{|x-2y|}{|x-2y|^2}}\right)$$ $$=$$ $$\lim_{(x,y) \to (2y,y)} \exp\left({\frac{1}{|x-2y|}}\right) \equiv \lim_{z \to 0} \exp\left(\frac{1}{|z|} \right) = \infty$$

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We have that by $t=x-2y \to 0$ we reduce to the simpler

$$\large e^{\frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}\to e^\infty=\infty$$

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