How can there be two different answers from two identical equations for the same value of $x$? $(x-1)/(x^2-1)$ and $1/(x+1)$. Given the following equation:
$$f(x) = \frac{x-1}{x^2-1}$$
find the limit of $f(x)$ when $x \to 1$.
I know how to solve it, by simplifying the above equation to $f(x)=1/(x+1)$, giving the answer $1/2$.
But my question is: $(x-1)/(x^2-1)$ is identical to $1/x+1$. How come $(x-1)/(x^2-1)$ gives $0/0$ when $x=1$, but $1/(x+1)$ gives $1/2$ when $x=1$?
 A: The graphs for $f(x)=(x-1)/(x^2-1)$ and $g(x)=1/(x+1)$ are not equal: they differ for exactly one value of $x$, namely the one you gave! Specifically, $f$'s graph is just that of $g$, but with a "hole" punctured in it at the point $(1, 1/2)$. It's a removable discontinuity, like this one:

The reason this happens is that $f(x)$ is just $g(x)$ multiplied by $(x-1)/(x-1)$. We can examine the behaviour of that function more closely:


*

*If $x \neq 1$, then $(x-1)/(x-1)$ equals 1.

*If $x = 1$, it is $0/0$, which is undefined. (It's an indeterminate form.)


So for all values of $x$ but one, we multiply by $1$, and the value of $f(x)$ and $g(x)$ is equal. Only for $x=1$ our new function gets a "hole" added to it.
A: There are no two answers! As it is, the function is not well defined for x=1, as division by 0 is not defined. That is, you can't just substitute x=1 in the original expression. It makes sense only as a limit. Else, once you identify it with 1/x+1 you can ascribe it a finite value at x=1.
A: Observe that the simplification $$\frac{x-1}{x^2-1}=\frac{1}{x+1}$$
is true iff $x^2-1\neq 0$ since this kind of manipulation is due to the properties of the real numbers and the quotient is not defined for $x^2-1=0$.
A: What you have in the first case is a limit of a quotient with an indefinite form.  
That is $\lim\limits_{x\to c} \dfrac{f(x)}{g(x)}$, where $\lim\limits_{x\to c} f(x)=0$ and $\lim\limits_{x\to c} g(x)=0$.  
Specifically you have $f(x)=x-1, g(x) = x^2-1$.   Both vanish at the limit point, $x=0$.
Such limits can be assigned a definite value using various techniques.   One such method is to cancel common terms which do not equal zero near the limit point.   (This term may equal zero at the limit point, as long as it does not do so on the approach.)
The term you cancelled was $(x-1)$.

So although $\frac{x-1}{x^2-1}$ and $\frac{1}{x+1}$ are only equal everywhere but at the point $x=1$, where the first function is discontinuous, both do approach the same limit there.
A: They are two different functions, but they differ only in their domains -- what x is allowed to be. And the forbidden value is x = 1.
You are, however, allowed to as closely with this forbidden fruit as you like. The closer you get to x = 1 in your initial equation, the closer you'll get to the value you're looking for.
Suppose you had a roadway with a gorge in its path. You can't travel the road to the other side because of the gorge. But you can build a bridge. That's what factoring did for you. The functions both go to the exact same place except where the gorge is.
A: "But my question is: $(x−1)/(x^2−1)$ is identical to 1/x+1. How come $(x−1)/(x^2−1)$ gives 0/0 when x=1, but 1/(x+1) gives 1/2 when x=1?"
They are not identical. As others have noted, they are equal for all values except for x = 1, where they differ.
Recall that you can never divide by zero. When reducing rational expressions in the way that you're doing here, you technically have to keep track of any values that would have caused you to divide by zero, because they're not in the domain of the original expression. This is a detail that is glossed over in many classes or testing contexts, but is very important to avoid confusion in exactly this way. In your particular case, cancelling the $(x - 1)$ factor in the numerator & denominator does not hold if $x = 1$, because that would be division by zero. Thus:
$(x-1)/(x^2-1) = 1/(x+1)$ for all values of $x$ is a false statement.
$(x-1)/(x^2-1) = 1/(x+1)$ for all values $x ≠ 1$  is a true statement.
