can we separate out absolute value function like this?

Question

$$ \lim_{x\to 2^-}\frac{x(x-2)}{|(x+1)(x-2)|}= \lim_{x\to 2^-}\left(\frac{x}{|x+1|}\cdot \frac{x-2}{|x-2|}\right) $$

So as the title says, is it okay to separate function under absolute value like this (i.e In form of Products) as shown in the denominator?

Answer 1

Yes, $|ab|=|a||b|$ holds for all $a, b\in\mathbb{R}$.

Answer 2

Yes, in general $|ab|=|a||b|$. You can check all the sign combinations to justify it.

Answer 3

Yes of course we can di that since the two expressions are equivalent, moreover in that case we can also go beyond that indeed

$$\lim_{x\to 2^-}\frac{x(x-2)}{|(x+1)(x-2)|}= \lim_{x\to 2^-}\left(\frac{x}{|x+1|}\cdot \frac{x-2}{|x-2|}\right)=\lim_{x\to 2^-}\left(\frac{x}{|x+1|}\right)\cdot \lim_{x\to 2^-}\left(\frac{x-2}{|x-2|}\right)$$

since the two limits exist and are finite.