Right now, I'm doing a question:

$$\lim_{(x,y)\to(1,0)}\frac{xy-y}{(x-1)^2 +y^2}$$

I know the limit doesn't exist, but I can't figure out how to prove it. I tried putting $x=1$, and getting $0/y^2$, and put $y=0$, got $0/(x^2-2x+1)$, but I don't think that does it.

(edit: this is not a duplicate; I'm having a hard time getting good explanations, hence why I asked)

  • 2
    $\begingroup$ Set $z=x-1$. Then this is the same as showing that the limit $\lim_{(z,y)\to(0,0)}\frac{zy}{z^2+y^2}$ does not exist. Then this is a very common example. See, e.g., math.stackexchange.com/questions/518357/… $\endgroup$ – Luiz Cordeiro May 25 '16 at 22:51
  • $\begingroup$ Parametrize lines a varying slope through $(1,0)$ then restrict the function to the lines to get one variable limits. Evaluate and see the limit depends on the slope of the line, hence it doesn't tend towards a single value, so it doesn't exist. $\endgroup$ – Charlie Frohman May 25 '16 at 22:51

Along the path $y=x-1$, we have

$$\begin{align} \lim_{(x,y)\to (1,0)}\frac{xy-y}{(x-1)^2+y^2}&=\lim_{(x,y)\to (1,0)}\frac{(x-1)^2}{2(x-1)^2}\\\\ &=\frac12 \end{align}$$

Along the path $y=0$, we have

$$\lim_{(x,y)\to (1,0)}\frac{xy-y}{(x-1)^2+y^2}=0$$

  • $\begingroup$ how did you chose to make y=x-1 instead of y=0 (because isn't y going to zero?) $\endgroup$ – KrissyMichaelsson May 25 '16 at 22:52
  • $\begingroup$ Indeed, $y\to 0$. Inasmuch as $x\to 1$, then $y=x-1\to 0$ $\endgroup$ – Mark Viola May 25 '16 at 22:53
  • $\begingroup$ oh, so you just related the values of x and y, because y is 1 less than x, hence y=x-1? $\endgroup$ – KrissyMichaelsson May 25 '16 at 23:01
  • $\begingroup$ Not quite. I merely found two paths on which the limit differs. Another way to see this is to let $x=1+r\cos(\phi)$ and $y=r\sin(\phi)$. Then, $$\frac{xy-y}{(x-1)^2+y^2}=\frac12\sin(2\phi)$$ $\endgroup$ – Mark Viola May 25 '16 at 23:04
  • $\begingroup$ I have a test Friday, so would whats in the previous comment have worked, and could it work in other problems (like limit as x,y approach 2, 4) of some function? $\endgroup$ – KrissyMichaelsson May 25 '16 at 23:05

Take la sequence of points $P_n=(1\pm \frac 1n,\frac 1n)$. You have $$\lim_{x\to \infty}\frac{(1\pm \frac 1n)\frac 1n-\frac 1n}{(\pm\frac 1n)^2+((\frac 1n)^2}=\pm \frac 12$$ These two distinct limits say that the limit doesn't exist.


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.