Proof: floor function is not differentiable at 0 Let $f(x)=[x]$ , show that $f(x)$ is not differentiable at $c=0$, 
 Here's my work so far:
1.$\frac {df(c)}{dx}=\lim_{x \to c}\frac {f(x)-f(c)}{x-c}$, $c=0$
2.$\frac {df(0)}{dx}=\lim_{x \to 0}\frac {f(x)-f(0)}{x-0}$, since $f(0)=0$
3.$\frac {df(0)}{dx}=\lim_{x \to 0}\frac {f(x)}{x}$
4.$\frac {df(0)}{dx}=\lim_{x \to 0}\frac {[x]}{x}$, use one-sided limit
5.$\frac {df(0)}{dx}=\lim_{x \to 0^+}\frac {[x]}{x}=\lim_{x \to 0^+}\frac {0}{x}=0$
6.$\frac {df(0)}{dx}=\lim_{x \to 0^-}\frac {[x]}{x}=\lim_{x \to 0^-}\frac {-1}{x}= \infty$
Since one-sided limits are not equal, therefore
$\lim_{x \to 0}\frac {[x]}{x}$  doesn't  exist, and the function is not differentiable at $c=0$.
Q1\ Is this approach correct for proving? If there are any mistakes please let me know.
Q2\ at step 6. , The limit from left side as c approaches 0 is equal to infinity, which is not what it shows on a graphing calculator(see attached picture). instead according to the graph, limit of the derivative from left side is 0, and it actually makes more sense to me since the function $f(x)$ on interval $0>x≥-1$ is just a horizontal line and the derivative should be $0$ around $x=0$ and not differentiable at $x=0$ itself, is the graphing calculator wrong or my calculation are wrong?
Please note this is not a homework problem.
http://i.stack.imgur.com/HtRfM.jpg
 A: Your reasoning is correct.
What the calculator shows is also correct. Indeed, the derivative at $x=-\epsilon$, where $\epsilon$ is a very small positive number, is $\lim_{x\rightarrow \epsilon} \frac{f(x)-f(\epsilon)}{x-\epsilon} = \lim_{x\rightarrow \epsilon}\frac{-1 - (-1)}{x-\epsilon} = \lim_{x\rightarrow \epsilon}\frac{0}{x-\epsilon} = 0$.
The fact that the left derivative at $x=0$ is $\lim_{x\rightarrow 0^-} -\frac{1}{x}$ (and, therefore, infinite, as you said) does not imply that the value of the derivative behaves like $-\frac{1}{x}$ as $x$ approaches $0$. 
A: Before testing for the equality of the two one-sided derivatives, should you not first check whether $f(x)$ is continuous at $x = 0$?  
The blue graph in the image is the graph of $f'(x)$, not of $- 1 / x$, for $x \in (- 1, 0)$.  
A: (1) You could just say "since $f$ is not continuous at $c=0$, then $f$ is not differentiable at $c=0$".
(2) Both graphing calculator and your calculation are correct. In some sense, the derivative of a function measures how much this function varies. Take a neighborhood of $c=0$, say $(-\epsilon, \epsilon)$ for an arbitrarily small $\epsilon >0$ and note that the variation of $f$ in this neighborhood is always $1$. Therefore, the rate of change is as bigger as you want.
Another way to see this is the following: define a sequence $(f_n)_{n\geq 2}$ of functions on $(-1,1)$ by
$$f_n(x)=
\left\{ \begin{array}{ll}
-1 & x\in (-1,-\frac1n] \\
\frac12 (nx-1) & x\in (-\frac1n,\frac1n)\\
0 & x\in [\frac1n,1)
\end{array}\right.$$
Note that
$\lim\limits_{n\to\infty} f_n(x) = f(x)$ and that the rate of change of each $f_n$ is $n$ (on $[0,1/n]$), so it's grow up as much as you want. Maybe, look at the derivatives
$$f'_n(x)=\left\{\begin{array}{ll}
0 & x\in (-1,-\frac1n]\cup[\frac1n,1) \\
\frac12n & x\in (-\frac1n,\frac1n)
\end{array}\right. $$
and think about $f'(x)$ as $\lim\limits_{n\to\infty} f'_n(x)$.
Ps.: Thinking about $f'$ like this isn't rigorous enough.
