Showing that the limit exists. I don't understand the solution of the following question: 

They are saying that when y = 0, F(x,y) = 0 and therefore it is continuous.
Does that really say that it is continuous? And how so? And also, they are saying that when y = 0, F(x,y) = 0 for all x belonging to all real numbers. But how can x be 0? Wouldn't that be 0*0/(0+0)? 
And then they are saying that when y does not equal 0, x is defined everywhere. And hence this function is continuous. 
I'm so confused about this question, I now it is not a hard question, but the solution does not make sense. 
Thank you!
 A: Let $f(x)=F(x,0)$.  Then, from the definition of $F(x,y)$ given in the OP we can write

$$f(x)=\begin{cases}0&,x\ne 0\\\\0&,x=0\end{cases}$$

Inasmuch as $f(x)\equiv 0$ for all $x$, it is a continuous function.
Similarly, let $g(y)=F(0,y)$.  Then, from the definition of $F(x,y)$ given in the OP we can write

$$g(y)=\begin{cases}0&,y\ne 0\\\\0&,y=0\end{cases}$$

Inasmuch as $g(y)\equiv 0$ for all $y$, it is a continuous function.
However, the function $h(x)=F(x,x)$ is given by 

$$h(x)=\begin{cases}\frac12&,x\ne=0\\\\0&,x=0\end{cases}$$

is evidently discontinuous at $x=0$.

We conclude that $F(x,y)$ cannot be continuous due to the discontinuity at the origin.


NOTE:
To make all of the preceding more concise, we simply note that $\lim_{(x,y)\to (0,0)}F(x,y)$ fails to exist (and hence $F(x,y)$ cannot be continuous at $(0,0)$) since on the path $x=t$, $y=0$ we have
$$\lim_{t\to 0}F(t,0)=0$$
while on the path $x=y=t$ we have
$$\lim_{t\to 0}F(t,t)=\frac12$$
A: At that point in the argument, they are specifically making no claim about continuity across $y$ values; only when moving across $x$ at any constant $y$.
They are asserting that $F(x\times 0)$ is continuous across all $x$, and that $F(x\times c)$ is continuous across all $x$ for non-zero $c$.   Therefore $F(x\times c)$ is continuous across all $x$ for an arbitrary constant $c$, zero or otherwise.

Also notice that the definition of $F$ is a piecewise function.   It is $0$ when both $x$ and $y$ are zero and otherwise it is $xy/(x^2+y^2)$.
Thus $F(x\times 0)=\begin{cases} 0 &:& x\times 0=0\times 0 \\ 0(x/(x^2+0^2)) &:&\text{otherwise}\end{cases} \\~=~ 0$

They then show the same result for keep in $x$ constant.   However, a different result arises on the part of $x=y$, as there is a discontinuity in 
$$F(x\times x) = \begin{cases}0 & : & x=0\\ \tfrac 12 & :& x\neq 0\end{cases}$$ 
A: The book is trying to show that
$$g(x) := \frac{xy}{x^2+y^2}1_{\{0\}}(x)$$ is continuous (on $\mathbb R$)
Regardless of the value of y.
$y$ here is treated as constant, but its value leads to very different $g(x)$'s. We have to show that in the different $g(x)$'s, we have a continuous (on R) function.

It's like if I asked you if this is continuous:
$$h(x) = x^21_{(0,\infty)}(x) + b1_{\{0\}}(x) + -x1_{(-\infty,0)}(x)$$
Case 1: $b=0$
Then $h(x)$ is continuous
Case 2: $b \ne 0$
Then $h(x)$ is not continuous
Since $h(x)$ is not continuous (on R) in each case ie regardless of the value of b, it is not continuous (on R).

Going back:
Case 1: $y=0$
Then $g(x) \equiv 0$.
Here, g is constant, and constant functions are continuous
(on R).
Case 2: $y \ne 0$
Then $g(x) = \frac{xy}{x^2+y^2}$ is defined on all $\mathbb R$*
Here, g is a rational function.
In single variable calculus, we are given that rational functions are continuous wherever they are defined (or some classes prove it. Whatever).
Since $g(x)$ is a rational function defined on $\mathbb R$, $g(x)$ is continuous (on $\mathbb R$).

*Note that if $y$ can be $0$, then this is not true.
$\frac{1}{x^2+b}$ isn't defined on $\mathbb R$ if $b \le 0$ because we may have $x=\pm \sqrt{b}$ but is defined on all $\mathbb R$ if $b>0$ since the denominator will never be zero.
Thus, $\frac{1}{x^2+b}$ is not continuous on $\mathbb R$ if $b \le 0$ because it is not defined on $\mathbb R$. However, $\frac{1}{x^2+b}$ is continuous on wherever it is defined for any $b$.
