# Two-Path Test to prove the limit does not exist

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".

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

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.