Left Limit of |x| at x=0: Conceptual Reason

[ktpr2]

This pretty book here says that \lim_{x\rightarrow 0^-}\frac {|x|}{x} is equal to \lim_{x\rightarrow 0^-}\frac {-x}{x} ...

I understand that we're taking a left sided limit, as x approaches 0, so x will always be negative, but we're putting that value in the absolute value function which should always return a postive result. So what's the conceptual reason for the absolute value function returning a negative number? The only thing I could come up with is the order of operation application is different; you take the absolute value of x first and then let that x approach 0 from the left, which would be result in a negative x.

 

[dextercioby]

It doesn't.It returns a positive value ALWAYS...Even for complex #...

Daniel.

 

[ktpr2]

uhhhhh... then can you prove that \lim_{x\rightarrow 0}\frac {|x|}{x} exists? :) My text, second edition Calculus concepts an contexts by james stewart, says it doesn't exist.

 

[dextercioby]

You mean the left or the right limit...?The partial limits are - and + infinity respectively...So the limit -------->0 (full) DOES NOT EXIST...Neither in Rbar.

Daniel.

 

[ktpr2]

wiat a second, x can be negative; its the y value, the range, that is always positive. But even still it appears that they're abusing the || function because when you slap a function on a number you use the result, the range, not the input (otherwise there's no point in using the functino in the first place). For any of you that have this text, its on page 115.

 

[dextercioby]

No,the limit is generally COMPUTED OVER THE DOMAIN OF DEFINITION.

Daniel.

 

[ktpr2]

says here the partial limits are +/- 1.

If it said \lim_{x\rightarrow 0^-}\frac {|x|}{x} = \lim_{x\rightarrow 0^-}\frac {x}{-x} = -1 i'd by that, but the text says \lim_{x\rightarrow 0^-}\frac {|x|}{x} = \lim_{x\rightarrow 0^-}\frac {-x}{x}

 

[Hurkyl]

You're asking why |x| = -x when x is negative? Isn't that part of its definition?

 

[ktpr2]

yeah that makes sense, but if x is negative then why isn't the denominator negative as well, since it's x? x-->0- holds for both variables in the function |x|/x

 

[Hurkyl]

x is negative when x < 0.

 

[dextercioby]

says here the partial limits are +/- 1.

If it said \lim_{x\rightarrow 0^-}\frac {|x|}{x} = \lim_{x\rightarrow 0^-}\frac {x}{-x} = -1 i'd by that, but the text says \lim_{x\rightarrow 0^-}\frac {|x|}{x} = \lim_{x\rightarrow 0^-}\frac {-x}{x}

That's because YOU EDITED YOUR POSTS.You had a square in the denominator...

Daniel.

 

[NateTG]

yeah that makes sense, but if x is negative then why isn't the denominator negative as well, since it's x? x-->0- holds for both variables in the function |x|/x

Let's say that
x=-5
or something like that.
Then we have
\frac{|x|}{x}=\frac{|(-5)|}{(-5)}=\frac{5}{-5}=\frac{-(-5)}{(-5)}=\frac{-x}{x}

Since x is negative, the negation in the numerator makes the numerator positive, and the denominator is negative.

 

[ktpr2]

dextercoiby - yeah that was a typo i removed; changes everything sorry about that.

hurkyl sure -x is negative but how does that force |x|/x to be negative. If x is negative, i'd expect something like -x/-x beause 2) |x| is -x and for x we're taking x -> 0- for denominator x.

 

[ktpr2]

okay. The way I'm looking at this process now is:

|x| gives us -x
---
x gives us plain old x

then we actually take the limit
and we get
-(-x)
----
- x

which gives us

-x/x = -1

is that's what going on?
[edited for conceptual misunderstandings]

 

[Bartholomew]

Just because x is negative doesn't mean you should put a negative sign in front of it.

 

[NateTG]

dextercoiby - yeah that was a typo i removed; changes everything sorry about that.

hurkyl sure -x is negative but how does that force |x|/x to be negative. If x is negative, i'd expect something like -x/-x beause 2) |x| is -x and for x we're taking x -> 0- for denominator x.

You seem to be confused about the following fact:

If x is negative, -x is not negative.

Take a moment and look at the sentence I just wrote, and make sure you understand it. You appear to be confusing negation (an operation) with something being negative (a property).

 

[ktpr2]

okay i got it now. thanks for all the help Peoples of Physics and Chemistry. (POPAC)

 

[ktpr2]

hmm so -x is negation of x, not negative x, which is the negation of positive x?
eidt - accidentally typed negative when ishouldve used negation

 

[NateTG]

hmm so -x is negation of x, not negative x, which is the negation of positive x?
eidt - accidentally typed negative when ishouldve used negation

One common misconception is essentially that
-x
is thought of as what would be written as:
-|x|

If that makes any sense at all.

 

[HallsofIvy]

If x is negative then, yes, |x|= -x when is positive because x itself is a negative number. \frac{|x|}{x}= \frac{-x}{x} which is a positive number over a negative number. That's negative.

 

[Galileo]

ktpr2, it's really not as complicated as it may appear from this thread.

The definition of |x| is:

<br /> |x|=\left\{<br /> \begin{array}{cc}<br /> x, &amp; \mbox{if} x \geq 0\\<br /> -x, &amp; \mbox{if} x&lt;0<br /> \end{array}\right.

So it's always positive (or zero).
Alternatively: |x| is equal to the distance from the origin and distances are always non-negative.

So |x|/x is positive if x is positive, because you are dividing something positive by something positive.
|x|/x is negative if x is negative, because you are dividing something positive by something negative.

Using the definition, |x|/x is equal to 1 if x is positive and -1 if x is negative.