Is $\frac{1}{ \frac{1}{x-1}}$ equivalent to $x-1$ even when $x=1$?
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Do you mean even when $x=1$? – Mar 26 '15 at 21:09
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@math Yes. my bad. – YoTengoUnLCD Mar 26 '15 at 21:22
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Wait, but doesn't $x = 1$ lead to division by zero in this case? – Robert Soupe Mar 27 '15 at 00:06
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$$\frac{1}{(\frac{1}{x-1})} = \big(\frac{1}{x-1}\big)^{-1} = \big((x-1)^{-1}\big)^{-1} = (x-1)^{(-1)\cdot(-1)} = (x-1)^1 = x-1,\ \ \forall x\neq1$$
The only thing you need to worry about is when $x=1$, in which case the original function is undefined, and all manipulations there after will also be undefined at $x=1$. For convenience, you may choose to define that point as $0$, but this choice must be explicitly stated.
jameselmore
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