# What is wrong with this limit argument?

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 $$x \leq 1$$?

• $\frac{x-1}{|x-1|} = -1$ for $x<1$. Apr 2, 2023 at 1:22
• 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. Apr 2, 2023 at 1:23

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.

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).

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.