29–34. Nonexistence of limits Use the Two-Path Test to prove that the following limits do not exist. $$\lim_{(x,y)\to(0,0)}\frac y{\sqrt{x^2-y^2}}$$

Observe that along the line $y=0$, $\lim_{(x,y)\to(0,0)}\frac y{\sqrt{x^2-y^2}}=\lim_{x\to0}\frac0{|x|}=0$, whereas along the ray $x=2y,y>0$, $\lim_{(x,y)\to(0,0)}\frac y{\sqrt{x^2-y^2}}=\lim_{y\to0}\frac y{\sqrt3y}=\frac1{\sqrt3}$.

Could anybody explain to me the meaning of the last phrase with $\frac1{\sqrt3}$? I understand the previous step and that the limit does not exist. But I don't understand the second approach to the limit and why it's called "ray".

  • 2
    $\begingroup$ A ray is just a half-line. $x=2y$ is a line and $x=2y, y>0$ is a ray. $\endgroup$ Commented Feb 26, 2023 at 7:17

1 Answer 1


In the first path you can approach from either direction ($y$ tending to $0$ from either positive or negative side) and you will get the same result.

The limit in the second path, however, only has the value $\frac1{\sqrt3}$ when approached from the positive direction in $x$ and $y$, i.e. it is a one-sided limit. This half-infinite path is called a ray.

After subbing $x=2y$ in the limit we have $$\lim_{(x,y)\to(0,0)}\frac y{\sqrt{(2y)^2-y^2}}=\lim_{(x,y)\to(0,0)}\frac y{\sqrt{3y^2}}=^*\lim_{(x,y)\to(0,0)}\frac y{y\sqrt{3}}=\frac1{\sqrt3}$$ where the starred equals sign relies on $y$ being positive.

  • $\begingroup$ Thanks for your help. But I am still unclear on the numbers. Why x=2y? Where does the sqrt(3)y come from? Because if I replace the x with 2y inside the root, I don't get the same value. Thanks in advance. $\endgroup$
    – Daddy
    Commented Feb 26, 2023 at 17:35
  • $\begingroup$ @AlejandroAlejandro See edit. $\endgroup$ Commented Feb 26, 2023 at 17:47

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