Is $(x+1)/(x^2-1)$ defined for $x=-1$? It might sound like a silly question, but I can't come up with a clear answer.
By looking at the expression, the answer should be "no", since $(-1)^2=1$ and we're in trouble. 
However, if I factorize: $(x^2-1) =(x+1)(x-1)$, $x=-1$ is still illegal. But now the terms cancel out and I am left with $1/(x-1)$, which is clearly defined for $x=-1$?
What happened? Was some information lost in the manipulation or was the original expression an "illusion"? I hope my question makes sense.

One way to look at it is to realize $\frac {ab}{ac} = \frac{b}{c}$ only if $a \ne 0$  If $a = 0$ then we can not say $\frac{ab}{ac} = \frac{b}{c}$.  Of course we can not say $\frac{ab}{ac}$ equals anything.
So we can not say $\frac{x+1}{x^2 - 1} = \frac 1{x - 1}$.  We can say  $\frac{x+1}{x^2 - 1} =\begin{cases} \frac 1{x - 1}; x\ne -1;\text{(Note:}\frac 1{x - 1}\text{ is undefined if } x =1\text{)} \\\text{undefined}; x= -1 \end{cases}  $
This is a subtly different statement altogether.
