Differentiability of $x/x$ A disagreement in my calculus class has arisen as to whether $f(x) = \frac{x}{x}$ is differentiable for the domain of all real numbers, including $0$.
According to our textbook, for a function to be differentiable at $a$, it must be continuous at $a$.  What we're not sure about is:


*

*Is $f(x)$ actually discontinuous at $0$?

*If it's discontinuous at $0$, how come we can still obtain a derivative of $0$ by hand using limits?  Mathematica seems to agree with $f'(x) = 0$.


Edit: To clarify, the specific function in question is $f(x) = \frac{x}{x}$
 A: This highly depends on how you are using the formula $\dfrac{x}{x}$ to define your function $f$. 
To be precise $f(x)=\dfrac{x}{x}=1$ when $x \not=0$ and is undefined when $x=0$. Because $f$ is not defined at $x=0$ it is not continuous there. Also, because it is not defined at $x=0$, its derivative is not defined at $x=0$. Consider the difference quotient: $\dfrac{f(0+h)-f(0)}{h}$. In this difference quotient "$f(0)$" is undefined (so you can't even begin to compute the limit).
That said, $x=0$ is a removable singularity. So we can continuously fill in the value $f(x)=1$ when $x=0$ and "repair" our function. Using the new formula: $f(x)=1$ (which now gives a function defined on the whole real line) we get a function differentiable everywhere. When Mathematica tells you that $f$ is differentiable at $x=0$, it is removing the singularity for you.
A: The real question, you need to answer first is:

How do you define $f$ at $0$?

Up to now, I don't see some $f$ defined at $0$. For a function to be continuous (let alone differentiable) at some point you need it to be defined at that point.
So:


*

*If you define $f(0)=1$, the function is constant, hence differentiable, hence continuous.

*If you define $f(0)$ different, the function will not be continuous, hence not differentiable.


What one may say in this situation is:
The function is not defined at $x=0$ but for neighborhoods $U$ of $x$, it is defined on $U\setminus\{x\}$, and the limit $\lim_{x\to 0}f(x)$ is well defined (i.e. independent of how $x$ approaches zero), hence

$f$ can be extended continuously to $x=0$.

A: The problem is your definition of $f$. If $f$ is the function defined by $f(x)=1$, then of course the derivative exists everywhere and is equal to zero.
If, however, you define the function $f$ as taking an input $x$ and dividing it by $x$ (i.e., $f(x)=x/x$), this function is technically undefined at zero (what does it mean to divide by $x=0$?). Therefore, you cannot talk about its derivative at zero, since one normally defines derivatives on open sets where the function is itself defined. However, you can show that $f^\prime(x)=0$ everywhere other than zero, and hence $f^\prime(x)\rightarrow 0$ as $x\rightarrow 0$.
A: The function isn't continuous at $0$. It's not discontinuous either. It just isn't, and that's the end of it. That being said, there exactly one nice, everywhere continuous and differentiable function defined on all of $\Bbb R$ that coincides with your $f$ on the domain of $f$.
Extending the domain of $f$ in this fashion is called removing a singularity, and isolated points where a given function is not defined, but where you can extend nicely in this fashion is called a removable singularity.
A: $$f(x)=\frac{x}{x}=\begin{cases}1 &, x \not=0 \\ \text{undefined} &, x=0\end{cases}$$
Yes, the function $f(x)=\frac{x}{x}$ is not defined at $x=0$. The right and left-hand limits of the function at $x=0^-$ and $x=0^+$ exist finitely and are equal to $1$. But $f(0)$ is undefined and hence you cannot make a proper mathematical statement involving $f(0)$, $\lim_\limits{x\to0^-}f(x)$ and $\lim_\limits{x\to0^+}f(x)$.
