The unknown in the derivative of the function We have function $f(x) = \lfloor x + \frac{1}{2}\rfloor x + x^2 $.   
If $ x\to \frac{1}{2}$ then $f'(\frac{1}{2}) = 1$ 
But If we use the definition this way instead:     

$\lim_{x\to \frac{1}{2}^-}{\frac{f(x) - f(\frac{1}{2})}{x - \frac{1}{2}}} = \lim_{x\to \frac{1}{2}^-}{\frac{x^2 - \frac{3}{4}}{x - \frac{1}{2}}} = \infty $ 
Which one is true and why? Is it possible to help me?
I am sorry for bad English. 
Thanks.
 A: Your function is not differentiable at $x=\frac{1}{2}$. Look at its graph and you will see what I mean:

Your function is discontinuous at this point, and so the rate of change of the function at that point is meaningless, unless you take the limit of the rate of change of the function up to that point. If you approach $x=\frac{1}{2}$ from the left, then the graph is basically that of 
$$y=x^2$$
and so the rate of change within the interval $(-\frac{1}{2}, \frac{1}{2})$ would be
$$y'=2x$$
and the limit of the rate of change as $x\to\frac{1}{2}$ from the left would be $1$. But if you approach from the right, you will be focusing on a steeper "segment" of the graph. Within that segment, the graph behaves like
$$y=x^2+x$$
and so the rate of change within the interval $(\frac{1}{2}, \frac{3}{2})$ is 
$$y'=2x+1$$
and the limit of the rate of change as $x\to\frac{1}{2}$ from the right would be $2$.
The answer is that neither of these are really "correct". The function is not differentiable at $x=\frac{1}{2}$, so its derivative at that point is meaningless.
