6
$\begingroup$

The limit is $$\lim\limits_{(x,y)\to(0,0)}\frac{\sqrt y}{\sqrt x}$$ I thought it was 1.

Also what about $$\lim\limits_{(x,y)\to(0,0)}\frac{ y}{ x}$$

$\endgroup$
  • $\begingroup$ undefined for negative $\endgroup$ – evil999man Apr 5 '14 at 7:53
  • $\begingroup$ Complex numbers exist then. $\endgroup$ – Superbus Apr 5 '14 at 7:55
11
$\begingroup$

Let $(y_n)=(\frac{1}{n})$ and $(x_n)=(\frac{1}{n^2})$. Then evaluate $\lim \limits_{n\to \infty}\frac{y_n}{x_n}$.

$\endgroup$
11
$\begingroup$

If any of the limits existed, then all the sublimits (for each case) would coincide. So take the limits along the parabolas $y=kx^2$ to find different sublimits.

$\endgroup$
  • $\begingroup$ Ah, thanks, that makes sense $\endgroup$ – Superbus Apr 5 '14 at 7:58
  • $\begingroup$ @LuciusTarquiniusSuperbus You're welcome. $\endgroup$ – Git Gud Apr 5 '14 at 8:00
  • $\begingroup$ Note that even if all the limits $y = kx$ coincide, they may still not exist (if you instead approach by $y = k x^2$...). This is a method to disprove, not prove. $\endgroup$ – Davidmh Apr 5 '14 at 15:30
  • $\begingroup$ @Davidmh My intent is to take the limit from the first quadrant. But there's no need to have such complications, so I changed my answer to consider parabolas. $\endgroup$ – Git Gud Apr 5 '14 at 17:35
  • $\begingroup$ What I meant is that, if from different paths you get different results, then it does not exist; but you cannot try all possible ways of getting there. There is at least one example where $y=kx$ converges, but $y=kx^2$ doesn't; and I am sure there are the oposite too... or cases where both converge but the limit still does not exist. The core of the issue is that you cannot try all possible ways of approaching your point. $\endgroup$ – Davidmh Apr 5 '14 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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