I'm having trouble understanding what happens when an algebraic expression in $\Bbb{R}$ that has a 1/x can be simplified into a form not containing 1/x.
For example, suppose I have:
$\frac{x^2-\frac{1}{x}}{x+\frac{1}{x}+1}$
If I simplify it in this way:
$\frac{\frac{x^3-1}{x}}{\frac{x^2+1+x}{x}}$ (Step 1)
$\frac{x^3-1}{x^2+1+x}$ (Step 2)
$\frac{(x-1)(x^2+1+x)}{x^2+1+x}$ (Step 3)
I finally get:
$x-1$ ,which is defined at $x=0$.
When I plug the expression $\frac{x^2-\frac{1}{x}}{x+\frac{1}{x}+1}$ in a graphing program it plots the line x-1 and WolframAlpha also simplifies this as x-1. But to do Step 1, are we not assuming that $\exists1/x\in\Bbb{R}$, which implies that $x\neq 0$?
Another example:
$x+\frac{1}{x+\frac{( 1+x)}{x}}$
$x+\frac{1}{\frac{x^2+ 1+x}{x}}$ (Step 1)
$\frac{x(\frac{x^2+ 1+x}{x})+1}{\frac{x^2+ 1+x}{x}}$ (Step 2)
$\frac{x^2+1+x+1}{\frac{x^2+ 1+x}{x}}$ (Step 3)
$\frac{x^2+x+2}{\frac{x^2+ 1+x}{x}}$ (Step 4)
And I finally get:
$\frac{x^3+x^2+2x}{x^2+x+1}$, which is defined at $x=0$. And again WolframAlpha gives this as a simplification and plotting software shows a curve that goes through the point $(0;0)$. But can Step 1 be done without assuming that $x\neq 0$??
This is sort of meta, but: These are examples of your proving that a given function defined on $\mathbb R{\setminus}\{0\}$ can be extended to a continuous function on $\mathbb R$. Formally, you are defining a new, different, function with a different domain, which happens to agree with the original function at all places where the original one is defined, and at $0$ as well, in such a way as to be continuous. The original expression assumes $x\ne 0$, and so do all the steps of your derivation, until the last step, where you say, "oh look, this formula makes sense for $x=0$ as well, and even looks better than the formula we started with." Yes, but this is an act of mathematical creation on your part: the end formula is a different one from the one you started with.