Allowing for jump discontinuites So I'm considering a definition given in one of my real analysis problem sets and trying to understand it a bit more intuitively.
DF: 
$$
\lim_{h\to 0}[f(x+h)-f(x-h)]=0
$$
My reasoning is that, this is essentially the same as continuity EXCEPT that we're allowing for jump discontinuities. Is this correct? I.e this definition allows for functions such as
$f(x)=x, \:\:x\in [-\infty,0) f(x)=x+b \:\: x\geq0$
If so, would it still be OK if x was undefined at $0$? EDIT:[SORRY THIS PART IS FALSE.]
Thanks
Ingvar
 A: The property 
$$\lim_{h\to 0}\, [f(x+h)−f(x−h)]=0 \tag{1}$$
is implied by continuity at $x$, but does not imply it. For example, any even function of $f$ satisfies at $x=0$, not matter how ugly it is. The Dirichlet function 
$f=\chi_{\mathbb Q}$ satisfies (1) at every rational $x$, though its discontinuities are far worse than jump discontinuity.
A: This definition tends be giving in the following problem in a few analysis books:

Suppose $f$ is a real function defined on $\mathbb{R}^1$ which satisfies
  $$                                                                            
  \lim_{h\to 0}\bigl[f(x + h) - f(x - h)\bigr] = 0                              
$$
  for every $x\in\mathbb{R}^1$. Does this imply $f$ is continuous?

As your question is written, I am not not sure what you are even asking about; however, if your post was spurred by the question I posed above, the answer is no.
Consider
$$                                                                            
  f(x) =                                                                        
  \begin{cases}                                                                 
    1, & x = 0\\                                                                
    0, & x\neq 0                                                                
  \end{cases}                                                                   
$$
Then $f$ is continuous at zero if for every $\epsilon > 0$ there exist a $\delta > 0$ such that $d_Y\bigl[f(x),f(0)\bigr] < \epsilon$ for all $x\in\mathbb{R}^1$ for which $0 < d_X(x,0) < \delta$. For $x\neq 0$, $0 < \lvert x - 0\rvert < \delta$. Then $\lvert f(x) - f(0)\rvert = 1$. Take $\epsilon = 1/2$ so $\lvert f(x) - f(0)\rvert > \epsilon$ and $f$ is not continuous at $x = 0$. For $x\neq 0$, $f(x) = 0$. Now for $h < \lvert x\rvert$, $x + h\neq 0\neq x - h$; therefore, $f(x + h) = f(x - h) = 0$.
$$                                                                            
  \lim_{h\to 0}\bigl[f(x + h) - f(x - h)\bigr] = 0                              
$$
For $x = 0$ and $h > 0$, $f(x + h) = f(h) = f(-h) = f(x - h)$ and again the limit is zero.
