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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.

Henrymerrild
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  • Notice how at $-1$, you have $0/0$, which is indeterminate, rather than undefined, which are different. – dbanet Sep 14 '16 at 22:00
  • @dbanet I'd say that from one (indeterminate) follows the other one (undefined...because it is indeterminate). The other way around isn't necessarily true, of course. – DonAntonio Sep 14 '16 at 22:02
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    "cancellation" is hidden division. To cancel the $(x-1)$ term you need to divide both numerator and denominator by $(x-1)$ and since division by $0$ is bad, you need to assume that $x\ne 1$ in order to do so. – Scott H. Sep 14 '16 at 22:27

6 Answers6

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There was some information lost - the division by $(x+1)$ is only valid when $x\ne-1$, since otherwise we are dividing by $0$. The function $\frac{x+1}{x^2-1}$ will have a hole at $x=-1$ (value does not exist), but is otherwise identical to $\frac{1}{x-1}$.

user2825632
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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.

fleablood
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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 $(x=-1)$.

Fimpellizzeri
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3

It is the exact same problem as saying:

Is $x/x$ defined for $x=0$ ?

The answer is no.

Even if you could change the formula with algebra to give it a value (which would obviously make sense), it is not defined per se.

E. Joseph
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The expression $\frac{x+1}{x^2-1}$ is undefined at $x=-1$ for the reason that you give: the denominator is $0$. And of course $\frac1{x-1}=-\frac12$ when $x=-1$. The explanation is that the two expressions are equal only where both are defined. It’s not true without qualification that

$$\frac{x+1}{x^2-1}=\frac1{x-1}\;;$$

the equality is valid only for those values of $x$ for which both expressions are defined. That excludes both $x=1$, where neither is defined, and $x=-1$, where the first is undefined.

Brian M. Scott
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Yes, it is not defined for the reasons you've stated above. To factorize, you divided by $(x+1)$, which is equal to zero, so it is an error.

John Doe
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