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.

This is what's called a removable singularity. In this case, we have that
$$\lim_{x\to -1}\frac{x+1}{x^2-1}$$
exists and is equal to $\lim_{x\to -1}\frac{1}{x-1}=-\frac12$, so setting the value of that function as $-\frac12$ yields a new function that is equal to the old one wherever the old one is defined $(\mathbb{R}\setminus\{-1\})$ and also continuous in the new point of definition $(x=-1)$.
