# How can I prove this limit doesn't exist?

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)

• 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/… – Luiz Cordeiro May 25 '16 at 22:51
• 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. – 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

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

• how did you chose to make y=x-1 instead of y=0 (because isn't y going to zero?) – KrissyMichaelsson May 25 '16 at 22:52
• Indeed, $y\to 0$. Inasmuch as $x\to 1$, then $y=x-1\to 0$ – Mark Viola May 25 '16 at 22:53
• oh, so you just related the values of x and y, because y is 1 less than x, hence y=x-1? – KrissyMichaelsson May 25 '16 at 23:01
• 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, $$\frac{xy-y}{(x-1)^2+y^2}=\frac12\sin(2\phi)$$ – Mark Viola May 25 '16 at 23:04
• 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? – 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.