- Thread starter bbm25
- Start date

- Joined
- Jan 29, 2005

- Messages
- 8,319

Please help me?

I think i understand: √x² = |x| Because the square of x when square-rooted could be negative or positive. (Is my understanding correct?)

I do not understand: |x| = -x

If \(\displaystyle \large x<0\) then \(\displaystyle \large |x|=-x\).

Now as \(\displaystyle x\to -\infty\) it is the case that \(\displaystyle x<0\).

- Joined
- Apr 15, 2019

- Messages
- 4

On the other hand, if the limit was approaching positive infinity, it would be the case that X>0, so you would have put X instead of -X. It is a case-by-case situation.This is exactly the same misunderstanding you had before.

If \(\displaystyle \large x<0\) then \(\displaystyle \large |x|=-x\).

Now as \(\displaystyle x\to -\infty\) it is the case that \(\displaystyle x<0\).

- Joined
- Dec 30, 2014

- Messages
- 3,652

Yes, it is a case by case situation. Just look at the definition of |x|. Sometimes |x|=x and other times |x| = -x. So which definition do you use? Well it depends of the value of x. For example |-4| = -(-4) = 4 while |14| = 14

- Joined
- Jan 29, 2005

- Messages
- 8,319

While that is true , it has nothing to do with what was asked.On the other hand, if the limit was approaching positive infinity, it would be the case that X>0, so you would have put X instead of -X. It is a case-by-case situation.

Okay, so to be absolutely clear here, what this now means is, given the above, the following therefore either isn't correct, or more accurately, can be 'overridden' if the question says so?

Yes, it is a case by case situation. Just look at the definition of |x|. Sometimes |x|=x and other times |x| = -x. So which definition do you use? Well it depends of the value of x. For example |-4| = -(-4) = 4 while |14| = 14

"Just for exactness, it is not always true that the absolute value of a number is always positive. The absolute value of a number (like 0) can be 0. The correct statement is that the absolute value of a number is never negative."

(From the earlier thread here https://www.freemathhelp.com/forum/threads/absolute-value-formula.115520/#post-451402)

- Joined
- Nov 12, 2017

- Messages
- 4,112

I think you are confusing the ideas of a

If x is negative, then -x, the negative of x, is positive. For example, if x = -3, then -x = -(-3) = 3, which is positive.

The absolute value of a number is either 0 (|0| = 0) or positive (|3| = 3 and |-3| = 3 are positive). This is true even when |x| is -x, because that happens only when x is negative.

- Joined
- Dec 30, 2014

- Messages
- 3,652

√x² = |x| since when you put in a positive number it comes out positive and if you plug in a negative you get back a positive number just like the absolute value function.View attachment 11776

Please help me?

I think i understand: √x² = |x| Because the square of x when square-rooted could be negative or positive. (Is my understanding correct?)

I do not understand: |x| = -x

- Joined
- Jan 29, 2005

- Messages
- 8,319

bbm25, this is my last try at helping you see this. You may need a live tutor to guide you step-by-step.

The whole point is to reduce \(\displaystyle \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{\sqrt {{x^2}} }}{x}} \right) = - 1\)

Thus we need to simplify \(\displaystyle {\frac{{\sqrt {{x^2}} }}{x}}\).

Well \(\displaystyle \sqrt {x^2}=|x|\), you said that you are O.K. with that.

How do we write \(\displaystyle |x|\) simply in terms of \(\displaystyle x\) 'alone' ? \(\displaystyle |x|=\begin{cases}x &: x\ge 0 \\ \Large -x &: x<0\end{cases}\)

You have asked repeatedly about that last point. Well once again: as \(\displaystyle x\to -\infty\) we have \(\displaystyle x<0\) so that \(\displaystyle |x|=-x\).

Once we have that then \(\displaystyle \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{|x|}}{x}} \right) = \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{ - x}}{x}} \right) = - 1\)

- Joined
- Nov 12, 2017

- Messages
- 4,112

My standard answer to "what happens if I do something other than what you teach?" is,

In this case, though, you might not see. The important thing here is that \(\displaystyle \sqrt{x^2}\) is not merely a power of x, so you would very likely do something wrong without realizing it. If, when you try it, you don't get the right answer, show us what you did so we can point out the error.

You don't generally

This was in response towards the book! The picture i took, not towards what you've been teaching me! Apologies!!!

- Joined
- Dec 30, 2014

- Messages
- 3,652

Fair question with a simple answer. While |x| sometimes equals x and other times equal -x it is the case that x ALWAYS equals x. So no need to replace x with -x (because they are NOT equal to one another)Why doesn't the x in the denominator (x + 1) also become -x?

- Joined
- Nov 12, 2017

- Messages
- 4,112

I don't think either of us took what you said as a criticism of whatThis was in response towards the book! The picture i took, not towards what you've been teaching me! Apologies!!!

So I

6) I will try to find the mistakes

9) i replace x with 0 instead of infinity because well.. that's what my book does

I also found it strange that my book replaces x with 0 instead of infinity but i found that if i do not do so, my answers wouldn't match the book's

Also, i notice that in this particular solution, even though they asked for the limit as x approaches negative infinity, they actually didn't even divide everything by the highest power in the denominator! Doesn't that make the answer wrong?

Last edited:

1) So for absolute clarity, this guy is wrong, and √x² does not = x and actually = |x|. Correct?

2) He also divides everything by the highest power in the denominator, which the book does not do. Is he also wrong here?