Why does $\lim\limits_{ (x,y) \to (0,0) } \frac{x-y}{\sqrt x - \sqrt y}$ not exist? Why does this limit not exist?
$$\lim\limits_{ (x,y) \to (0,0) } \frac{x-y}{\sqrt x - \sqrt y}$$
If you set y = 0, the limit goes to zero. If you set x = 0, the limit goes to zero.
You can also manipulate it with algebra to get zero.
However, if x=y you have zero/zero before you even evaluate the limit but is that proof enough? 
Thanks!
From Larson Calculus 13.2 Exercise 27
 A: The main problem with this exercise is the following: The expression
$$\Psi(x,y):={x-y\over\sqrt{x}-\sqrt{y}}$$
is undefined when $x<0$ or $y<0$, or $x=y$. In this situation one can argue in two ways:
(i) You can say that as a general rule the domain of definition of an expression is the set of $(x,y)$ for which it can be evaluated without asking supplementary questions. In the case at hand this is the set $$\Omega:=\{(x,y)\in{\mathbb R}^2\>|\>x\geq0,\  y\geq0, \ x\ne y\}\ .$$
The point $(0,0)$ is a limit point of $\Omega$, hence it makes sense to consider $\lim_{(x,y)\to(0,0)}\Psi(x,y)$. Now for all $(x,y)\in\Omega$ one has
$$\Psi(x,y)={x-y\over\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\ ,\tag{1}$$
and here the right hand side obviously converges to $0$ when $(x,y)\to(0,0)$.
(ii) You can say that the identity $(1)$ allows to extend the function defined by ${x-y\over\sqrt{x}-\sqrt{y}}$ continuously to all of $\bigl({\mathbb R}_{\geq0}\bigr)^2$. This is like defining ${\sin x\over x}$ to be $1$ at $x=0$. But this is a voluntary act, and is not stipulated in the formulation of the problem. If you want to adopt this position then the limit is of course again $=0$, since the function $(x,y)\mapsto\sqrt{x}+\sqrt{y}$ is continuous at $(0,0)$.
A: Well it is the same than $\displaystyle{\frac{x^2}{x}}$. 
Is it a $\frac 00$ limit or do we use algebra to simplify this to $x$ and say it is going to $0$ ?
In fact you can manipulate any expression to make it undefined if you want. But we make a postulate that for a function $f$ if after algebric manipulations the new function $\tilde f$ is well defined then we can extend $f$ by continuity to $\tilde f$.
In general we do not detail all of these considerations, we just assume the extension with the maximum domain fo definition.
In our case, we have 
$\begin{cases} 
\text{on }A=\{x\ge 0,\ y\ge 0,\ x\neq y\} & f(x,y)=\frac{x-y}{\sqrt x-\sqrt y} \\
\text{on }\bar A=\{x\ge 0,\ y\ge 0\} & \tilde f(x,y)=\sqrt x+\sqrt y
\end{cases}$
It is easy to see that :


*

*$f$ is continuous on $A$ as a composition of continuous functions

*$\tilde f$ is well defined and continuous on $\bar A$

*the restriction to $A$ of $f$ is $\tilde f:\qquad f_{/A}=\tilde f$


So we generally choose to work with $\tilde f$ defined on $\bar A$ instead of $f$ defined on $A$, so as to get rid of the somehow artificial singularity when $x=y$.

Now for the original question : $\tilde f(0,0)=0$ so the limit for $f$ does exists and we can choose to extend $f$ by continuity in this point by setting $f(0,0)=0$.



Edit: 
A little more into details about extending by continuity.
The theorem says : if a function $f$ is cauchy-continuous on a domain $A$ then it can be extended in a unique way to a function $\tilde f$ continuous on $\bar A$

For a Cauchy sequence in $A$ we have $f(x_n,y_n)=\tilde f(x_n,y_n)=\sqrt{x_n}+\sqrt{y_n}$ which is still cauchy according to the post forward for $x_n\ge 1$ and $y_n\ge 1$ : $\sqrt{a_n}$ is also cauchy if $a_n\ge 1$ is cauchy
For $[0,1]\times[0,1]$ this is not an issue either since this is a compact on which $\tilde f$ is continuous.
Finally the two remaining bands are out of interest, since $f=\tilde f$ already on these.
Thus $f$ is continuously extentable to $\bar A$ by $\tilde f$.

So it does not matter that it is a single point $x=0$ like in the case $\frac{x^2}{x}$ or a line $x=y$ like in the case of $f(x)$, what's matter is that in the restricted domain of definition the function is Cauchy-continuous (given that $(\mathbb R,|\cdot|)$ is complete, we can always conclude positively that the extension is sound).

Also a Cauchy sequence being bounded (let say inside a compact $K$), what's really important is not the behaviour at infinity but the continuity of the extension near the singularity. We want $\tilde f$ simply continuous over $\bar A\cap K$. ($\tilde f$ is then continuous over a compact, thus uniformly continuous, thus cauchy continuous). It is more constraining than just simple continuity of $f$ over $A$.

For instance the counter example given by Hanul Jeon is the above mentionned post, does not work because $f(x)=1/x,\ a_n=1/n$ then $\tilde f$ would not be continuous in $0$ no matter which value you assign to $\tilde f(0)$.
