Query on the ability to simplify the ratio of two functions taking in account that the function is incontinuous First of all, I will apologize for asking such a low-level question on this platform. My math teacher and I have had a disagreement on the procedure when encountering a certain problem. To give a recent example, the teacher gave us the following problem: "If $f(x) = x^2-1$ and $g(x) = x-1$, what is the value of $(\frac{f}{g})(x)$? My approach to this problem was to first recognize that both the dividend and divisor are explicitly stated to be functions. My thought was that since both are functions I couldn't do anything to simplify the function. My answer was $(\frac{f}{g})(x) = (\frac{x^2-1}{x-1})$. The teacher's approach was to say that you could factor $f(x)$ into $(x+1)(x-1)$ and then cancel out the $(x-1)$ in the denominator with the one in the numerator. Her final answer was $(\frac{f}{g})(x) = x+1$. I thought that this could not work because it is intrinsic to $(\frac{f}{g})$ that it is not defined at $x=1$ because that would cause the denominator to be zero. Thus, if the argument is that $(\frac{x^2-1}{x-1})$ equals $x+1$, it would require that they behave the same in every situation. If then one was completely continuous and the other not, wouldn't that imply that the simplification my teacher used was incorrect and resulted in a different function all together? That is my question and again my apologies for the inconvenience.
 A: Perhaps a couple of hundred years ago, your teacher's interpretation would have been regarded as correct, but now that Mathematics has been formalised, we understand functions somewhat differently, and nowadays, your interpretation is regarded as correct.
We now understand that a function is not just a thing that takes in an input and spits out an output: you need to say what is allowed to be an input to the function as well: we call this the domain of the function. For example, if we have a function $F$ defined by $F(x)=x+1$, it can act on real numbers, because we know how to add $1$ to a real number. It can't act on, say, $\text{table}$, because we have't a mathematically defined way of adding a number to an object like $\text{table}$. The domain of $F$ could be several other things: only the rational numbers, or only the integers, for example, but in basic Mathematics the reals are normally assumed to be the domain of a function if it is not specified.
In your case, it is indeed true that
$$ \frac{x^2-1}{x-1} = x+1 , $$
provided that $x \neq 0$. We need this condition because this identity comes from the uncontroversial polynomial identity
$$ x^2-1 = (x-1)(x+1), $$
by dividing by $x+1$, and we can't make sense of this if $x-1=0$. Therefore, $(f/g)(x)$ is not defined at $x=1$. It is defined elsewhere, and is equal to $x+1$ as you say. So we can write
$$ (f/g)(x) = x+1 \quad (x \neq 1) , $$
or more explicitly, $(f/g)(x) = x+1 $, the domain of $f/g$ being $\mathbb{R} \setminus \{1\}$.
If you want an authority to quote, you could not do much better than what G. H. Hardy has to say on this matter (this is not the first time I have quoted this on MSE, either):

The function $\frac{x^2-1}{x-1}$ has no value for $x=1$; for $x=1$, $\frac{x^2-1}{x-1}$ is strictly and absolutely meaningless. The fact that its limit for $x=1$ is $2$ is entirely irrelevant. The functions $\frac{x^2-1}{x-1}$ and $x+1$ are different functions. They are equal when $x$ is not equal to $1$. Similarly the function $y=\frac{x}{x}$ is $=1$ when $x \neq 0$ and undefined for $x=0$. To calculate $f(x)$ for $x=0$ we must put $x=0$ in the expression of $f(x)$ and perform the arithmetical operations which the form of the function prescribes, and this we cannot do in this case.

(Mathematical Gazette 1907, 4 pp. 13–14)
