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My question is very very basic but, for the life of me, I'm confused for whatever reason.

I know that if $x^2 = 5 $ then +$\sqrt{5}$ and $-\sqrt{5}$ are the solutions for $x$.

The reason, as I understand it, is that, in a function, $x$ can have two values while $y$ can only have one. And both values, when squared equal to $5$.

I also know that $\sqrt{4} = 2$ and $2$ only. The reason is we deal now with a square root function which only leads to positive values on the $y$ axis, otherwise, it wouldn't be a function as $y$ would have two values as in $\sqrt{x} = +y$ and $-y$, which isn't possible.

Now what about this then?

$y= \sqrt{x^2}$ ?

If I consider, for the sake of the example, that $x^2 = 4$ then we've already said that $\sqrt{4}$ only equals $2$, not $-2$. So the answer should be $x$ only, not both $x$ and $-x$. This makes sense somehow otherwise I would get two values for $y$ which is forbidden when it comes to functions.

So this leans towards the fact that $y= \sqrt{x^2}$ is definitely equal to $x$.

But then I see this on a youtube course:

enter image description here And this totally lost me, even though it's trivial.

I don't know why that second line is using the absolute value (probably to emphasize that it remains a positive $x$ which is consistent with what I concluded here above) but then the last line considers $-x$ as an eligible value this time and the part is beyond me. I know I'm overthinking it but I lost confidence with square roots right now. I need to get back to the basics.

Thanks for your patience and input.

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    $\begingroup$ When we take square roots, we really mean the so-called principal square root, when the result is positive. In general, $\sqrt{x} = |x|.$ $\endgroup$
    – Sean Roberson
    May 8, 2020 at 3:47
  • $\begingroup$ Ok thank you very much but then, how come it turns out as a -x on the third line of this limit, all of a sudden ? $\endgroup$
    – Bachir Messaouri
    May 8, 2020 at 3:55
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    $\begingroup$ Because $x\to-\infty$ says that we're looking at negative numbers $x$. $\endgroup$
    – Ted Shifrin
    May 8, 2020 at 4:10

1 Answer 1

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We say $\sqrt{x}$ refers to the principal root of $x$, which in the case of the positive real numbers refers to the root of $x$ that is greater than zero. So, $\sqrt{x^2}$ is the principal root of $x^2$. Now as you know, the square roots of $x^2$ are $x$ and $-x$, but only one of those is positive! Since the principal square root only cares about the positive root, it is the same as the absolute value in this instance.

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  • $\begingroup$ Ok thank you very much but then, how come it turns out as a -x on the third line of this limit, all of a sudden ? $\endgroup$
    – Bachir Messaouri
    May 8, 2020 at 3:54
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    $\begingroup$ @BachirMessaouri I guess what's happening here is that $|x|=-x$. Remember this happens whenever $x \le 0$. Since the limit is approaching negative infinity, $x \le 0$. $\endgroup$
    – healynr
    May 8, 2020 at 3:55
  • $\begingroup$ That's the part that is confusing me a lot. For 2 reasons. first because this means the square root of x^2 could lead to -x because it is an option eventually (proof on the third line), even though the scare root should prevent that. And second, I suspected it's because the limit leans towards - infinity but it should mean the value of x is negative not that the sign of the variable x is negative. Of course you are right, no question about it, I just fail to understand the logic. $\endgroup$
    – Bachir Messaouri
    May 8, 2020 at 4:00
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    $\begingroup$ Ok, got it after a few minutes of thinking ;-) Thank you. $\endgroup$
    – Bachir Messaouri
    May 8, 2020 at 4:14

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