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 $x\neq 0$??