What does this sentence mean? "$\lim_{x\to x_0}$ exists at every point $x_0$ in $(-1,1)$." 
What does this sentence mean?
  $$\lim_{x\to x_0} \;\text{exists at every point}\; x_0 \; \text{in} \; (-1,1).$$  

$(1,-1)$ is just an example point. The topic is finding whether  limit  functions are true or false and this is one of the questions. 
 A: The $(-1,1)$ notation is sometimes used to express open intervals. Alternatively $[-1,1]$ is used to express closed intervals.
So really it means that limit exists for $-1>x_0>1$ where as a closed interval $[-1,1]$ would include $-1$ and $1$; $-1 \geq x_0 \geq 1$ 
A: For every $x_0$ inside of the interval $(-1,1)$, meaning all points bigger than $-1$ but less than $1$, not including those two numbers, the limit of the function as $x\rightarrow x_0$ neither blows up, nor does it oscillate, it approaches some finite value. 
A: As the "contour example" take  $$f\left( x \right) =\ln { \left( 1-{ x }^{ 2 } \right)  } $$ it has limit in every point in $\left( -1,1 \right) $ and in  $\mathbb{R}\setminus \left( -1,1 \right) $ hasn't
A: The sentence in question means that on the open interval $(-1,1)$ on the x-axis, the function approaches some fixed value. This means that the function is connected when x is within this interval. Note, this does not mean that the function is continuous. Rather, only such that at every x value in the interval the function approaches a finite value.
$$\lim_{x\to x_0}$$ means that the condition below is true,
$$\lim_{x\to x^+_0} = \lim_{x\to x^-_0}$$
The above conditions basically mean that the right and left limits on the interval exist, which also means that there a value exists for every $x_0$ on this interval. The right and left limits being equal just means that the function approaches the same value from the left and right sides.
For example, this would satisfy your above statement
but this wouldn't (filled-in circles represent the actual function value while empty circles are holes within the function),

