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What is wrong in this argument?

$$ \lim_{x \rightarrow 1^-} \frac{x-1}{|x-1|} = \lim_{x \rightarrow 1^-} \frac{x-1}{\sqrt{(x-1)(x-1)}} = \lim_{x \rightarrow 1^-} \frac{\sqrt{x-1}}{\sqrt{x-1}} = 1$$

I know it is wrong because the book I’m working with shows that the limit should be $-1$ and not $1$. I suspect that it has something to do with the second equality, since I should really want $\sqrt{1-x}$ in the numerator because $x \leq 1$?

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    $\begingroup$ $\frac{x-1}{|x-1|} = -1$ for $x<1$. $\endgroup$
    – Stephen
    Apr 2, 2023 at 1:22
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    $\begingroup$ You seem to be writing $x-1$ as $\sqrt{x-1}\sqrt{x-1}$ in the numerator, which you cannot do when $x$ is smaller than $1$, and you cannot decompose the square root in the denominator into a product of square roots, because they are not defined when $x$ is smaller than $1$. So the same mistake twice. $\endgroup$ Apr 2, 2023 at 1:23

3 Answers 3

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You're right in thinking that the problem is with your second equation. On the right-hand side you have written $\sqrt{x - 1},$ which is not defined when $x < 1.$

If you work it so that you use only $\sqrt{1 - x},$ not $\sqrt{x - 1},$ you can solve the problem correctly.

But even simpler, for $x < 1$ you know that $\lvert x - 1\rvert = -(x - 1),$ therefore

$$ \lim_{x \rightarrow 1^-} \frac{x-1}{\lvert x - 1\rvert} = \lim_{x \rightarrow 1^-} \frac{x-1}{-(x-1)} = \lim_{x \rightarrow 1^-} -1 = -1. $$

No square roots are required.

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You make a jump between steps 2 and 3 but appear to be using the rule that $\sqrt{ab} = \sqrt{a}\sqrt{b}$. That rule is only valid when $a$ and $b$ are both positive (or one of them is zero).

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You can do that too. $\displaystyle \lim_{x \to 1^{-}} \dfrac{x-1}{|x-1|}= \displaystyle \lim_{x \to 1^{-}} -\dfrac{1-x}{\sqrt{(x-1)^2}}=\displaystyle \lim_{x \to 1^{-}}-\dfrac{1-x}{\sqrt{(1-x)^2}}=\displaystyle \lim_{x \to 1^{-}}-\dfrac{1-x}{|1-x|}=\displaystyle \lim_{x \to 1^{-}} -\dfrac{1-x}{1-x} = -1$ . Note that the factor $\sqrt{1-x}$ that you mentioned is simplifed or canceled out with the same factor at the denominator. Thus you don't have it in the numerator any more.

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