3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ hint $(x-2y)^2=|x-2y|^2$ $\endgroup$ – dmtri Dec 15 '18 at 8:36
  • 1
    $\begingroup$ $e^{\frac{1}{0}}=\infty$ if we go frrom the right of $0$ $\endgroup$ – dmtri Dec 15 '18 at 8:39
5
$\begingroup$

It is :

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

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.