Determining $\lim_{(x, y) \to (2y, y)} \exp(\frac{|x-2y|}{(x-2y)^2})$

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.

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

$$\lim_{(x,y) \to (2y,y)} \exp\left({\frac{|x-2y|}{(x-2y)^2}}\right) = \lim_{(x,y) \to (2y,y)} \exp\left({\frac{|x-2y|}{|x-2y|^2}}\right)$$ $$=$$ $$\lim_{(x,y) \to (2y,y)} \exp\left({\frac{1}{|x-2y|}}\right) \equiv \lim_{z \to 0} \exp\left(\frac{1}{|z|} \right) = \infty$$
We have that by $$t=x-2y \to 0$$ we reduce to the simpler
$$\large e^{\frac{|x-2y|}{(x-2y)^2}}=e^{{|t|}/{t^2}}=e^{{1}/{|t|}}\to e^\infty=\infty$$