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Is there any difference in the two functions, It is given that they are defined for all real $x$ except $x=0$

Is it somehow similar to the fact that functions $\sin x/\sin x$ and $1$ aren't the same?

Golden_Ratio
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amaydixit11
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    No, the two things aren't related, because $\frac x{\lvert x\rvert}$ and $\frac{\lvert x\rvert}x$ have the same domain. For $x\in\Bbb R\setminus{0}$, you can inspect the two cases directly and see that the two functions are equal. – Sassatelli Giulio Mar 09 '23 at 14:57
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    For what its worth, they are different if talking about complex numbers. $\frac{|i|}{i}=-i$ while $\frac{i}{|i|}=i$ for example. – JMoravitz Mar 09 '23 at 14:59
  • You are probably considering only real numbers; I'd use the definition of absolute value to work out the possible cases for each of your 2 expressions. – ryang Mar 10 '23 at 19:18

2 Answers2

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Assuming the functions you described are both maps from $\mathbb{R} - \{ 0 \}$ to $\mathbb{R}$, then they are the same. We can show this by considering an arbitrary $x \in \mathbb{R} - \{ 0 \}$. There are two cases: $x > 0$ or $x < 0$. In the former case, $|x| = x$, and so $\frac{|x|}{x} = \frac{x}{|x|} = \frac{x}{x} = 1$. In the latter case, $|x| = -x$, and so $\frac{|x|}{x} = -\frac{x}{x} = \frac{x}{|x|} = -1$.

K. Jiang
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Note for nonzero real $x$, we have $|x|^2=x^2, $ so

$${x \over |x|}={x \over |x|}\times {|x|^2 \over x^2}={|x| \over x}.$$

Golden_Ratio
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