Is $x/x$ equal to $1$ My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.
On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero. 
I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms. 

Note that this question arose after reading this answer and its first comment.
 A: The function $f(x) = \frac{x}{x}$ is defined for all $x \in \mathbb{R}\setminus\{0\}$. Its limits exist from the left and from the right at $0$, but it is not defined at $0$. It doesn't matter if you "simplify" first and then "check" or the other way around, because as you pointed out you're not allowed to divide by $0$. Thus 
$$f(x) = \begin{cases} 1 & x \neq 0 \\ \text{undefined} & x = 0\end{cases}.$$
Consider the same question,
$$g(x) = \frac{x^2 - 2x + 1}{x-1}.$$
What is this function? 
A: For any real number $x\neq0$, $$x=x,$$$$\Leftrightarrow$$$$\dfrac{x}{x}=1.$$
$x=0$ $\Rightarrow x=x,$ but now we cannot divide by $x$. (Why?)
In fact, this is an indeterminate form. 
Suppose that, $\dfrac{0}{0}=\color{red}{1}$. Hence, $0=\color{red}{1}\cdot0$. 
Therefore, $\color{red}{2}\cdot0=\color{red}{1}\cdot0$ since $\color{red}{2}\cdot0=0$.
So, $\color{red}{\dfrac{2}{1}}=\dfrac{0}{0}$. 
Thus $\color{red}{2}=1.$
A: Note: I'm assuming $x\in R$ in this whole post.
Indeed $x/x$ is only defined when $x\ne 0$. And wherever it is defined, its value is $1$.
However when people (especially in fields other than mathematics where mathematics is used) talk about such expressions, often what they really mean is: "The continuous (or, more generally, "well-behaved") function determined by the values of the given expression where that expression is defined". And there exists indeed exactly one continuous function $f:\mathbb R\to\mathbb R$ so that $f(x)=x/x$ for $x\ne 0$, and that is the constant function $f(x)=1$ for all $x\in\mathbb R$.
As a concrete example, consider the function $\sin(x)/x$ which appears in physics in the amplitude of a wave when describing diffraction on a slit. This expression is clearly not defined at $x=0$. However is is of course silly to assume that the amplitude of a wave is not defined at some point. What physicists actually mean is the continuous function
$$f(x)=\begin{cases}
\frac{\sin x }{x} & x\ne 0\\
1 & x=0
\end{cases}$$
But usually that function isn't written that way; the interpretation is implicitly assumed.
A: "On one hand the division should simplify to 1, ..."
No it does not: When you divide something by x (which is what you do in your simplification if you think about it) it is undefined for x=0. So when you simplify correctly you will also have the answer to your question.
A: As linked to in the post you link to, $\frac{0}{0}$ is an indeterminate form - that is, there is no answer at all. It's particularly curious because it's an example of both $\frac{x}{x}$ and $\frac{y}{x}$, the former of which equals $1$ when $x\neq0$ and the latter of which has a limit that tends to either positive or negative infinity when $y\neq0$ as $x$ tends to 0.
A: I disagree with all answers I've seen here, on the internet and even at MIT. What if you have $x^2 \le 4x^3 + 5x^2$. To solve you can divide both sides by $x^2$ which gives you $1 \le 4x + 5 \Rightarrow -4 \le 4x \Rightarrow -1 \le x$. But now since you divided by a power of $x$ to solve it you suddenly can't have $x = 0$ any more? It is inconsistent. I believe in simplifying instantaneously before division occurs.
