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Say, $f(x) = \dfrac{x-1}{x^2-1}$ and $g(x) = \dfrac{1}{x+1}$. Does $f = g$? I can reduce $f(x)$ to $g(x)$ but my textbook says they are not equal, can someone please explain to me why?

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    $\begingroup$ The domains are different: $ \ g(x) \ $ is undefined at $ \ x \ = \ -1 \ $ , while $ \ f(x) \ $ is undefined at both $ \ x \ = \ -1 \ $ and $ \ x \ = \ 1 \ $ . At all other values of $ \ x \ $ where both functions are defined, they will produce the same values. $\endgroup$ – colormegone May 4 '14 at 3:10
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In reducing $f$ to $g$, consider the steps you performed. Odds are, you factored out $(x-1)$ from the numerator and denominator of $f$ to give you

$$f(x) = \frac{(x-1)}{(x-1)} \frac{1}{x+1}$$

and then you concluded that the fraction on the left was equal to $1$ and so could be ignored. But that's not quite correct - there is a value of $x$ for which $(x-1)/(x-1)$ is not equal to $1$, namely $x = 1$. When $x = 1$, the fraction on the left is undefined, since the denominator is $1-1 = 0$, and you can't divide by zero.

This should suggest to you that you should see what happens when you evaluate $f$ and $g$ at $x = 1$. When we do that, we find that $g(1) = 1/2$, but that $f(1)$ is undefined (because of division by zero). Since $f$ and $g$ do not have the same value at all values of $x$, we conclude that $f$ and $g$ are not equal.

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  • $\begingroup$ I haven't considered the division by zero error, so I thought I could just cancel them out. Thanks. $\endgroup$ – mib1413456 May 4 '14 at 3:12
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    $\begingroup$ Adding to this: $f$ has what is called a removable singularity, informally a hole, at $x=1$. Note that this is not a removable discontinuity as it is sometimes inaccurately called. $\endgroup$ – Zubin Mukerjee May 4 '14 at 3:17

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