Finding the limit of the equation

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression \(\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}\). Participants are exploring the behavior of this limit as \(x\) approaches zero from both the positive and negative directions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss splitting the limit into two cases based on the direction of approach to zero. There are attempts to clarify the expressions used in the limit evaluation and to question the correctness of certain manipulations. Some participants raise concerns about the existence of the limit based on the differing behaviors from each side.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning. Some have suggested using the sequential criterion for limits to formalize the argument regarding the non-existence of the limit. There is an acknowledgment that the limits from the two sides do not converge to a common value.

Contextual Notes

Some participants express uncertainty about their understanding of limits, particularly in cases where factoring and substitution are involved. There is a mention of a lack of clarity in the professor's explanation during class, which contributes to the participants' hesitance in stating their conclusions about the limit.

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Homework Statement


Evaluate the limit if it exists

Homework Equations



[tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}[/tex]

The Attempt at a Solution



1) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|}[/tex]

2a) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{x}[/tex]

2b) [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{-x}[/tex]

3a) [tex]\lim_{x\to 0} 0 = 0[/tex]

3b) [tex]\lim_{x\to 0} \frac{1}{2x}[/tex]

4)[tex]\lim_{x\to 0} \frac{1}{2}x[/tex]

5)[tex](\frac{1}{2})0 = 0[/tex]

Did I do this correctly?
 
Last edited:
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shwanky said:
Did I do this correctly?

The limit is quite different if you approach zero from the negative direction or the positive direction. Split it into these two cases as I think you have tried to do, but 1a) 1b) 2a) 2b) etc is not the clearest way to express this. But finally 1/(2*x) is not equal to (1/2)*x. That's BAD.
 
Well, you didn't do it right as it doesn't exist.

1/x goes to -\infty as x->0 from the left. but -1/|x| is also going to -\infnty.

Use the sequence criterion of limits for a more formal proof.

Also note 1/x + 1/x is not 1/2x.
 
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Thanks, I kind of figured it was wrong. I'm still a little shakey on limits. If it's a problem that I can factor and substitute, I'm okay, but my professor didn't clearly explain this in class. I can usually see that the limit doesn't exist, but I don't know how to state it. Should I just take the + and - limits and show that they don't approach a common point on an open interval as ZioX did?
 
Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.
 
OMG! I can't believe it was that simple... :(
 
Define x_n=-1/n. This is a sequence converging to 0. But 1/x_n-1/|x_n|=-2n which doesn't converge to anything. Therefore 1/x-1/|x| has no limit as x tends to zero.

This is the sequencial criterion for limits. A function f converges to L as x converges to a iff for every sequence x_n converging to a has the property that f(x_n) converges to L.
 
For the limit to exist, [tex]\lim_{x\to 0} \frac{1}{x}-\frac{1}{|x|} =\lim_{x\to 0^{+}} \frac{1}{x}-\frac{1}{|x|} = \lim_{x\to 0^{-}} \frac{1}{x}-\frac{1}{|x|}[/tex]

However we can see that [tex]\lim_{x\to 0^{+}} \frac{1}{x}-\frac{1}{|x|}=0[/tex] but [tex]\lim_{x\to 0^{-}} \frac{1}{x}-\frac{1}{|x|} = \lim_{x\rightarrow 0} \frac{-2}{x}[/tex], which are obviously not the same.
 
[tex]\frac{1}{2x}\ne \frac{1}{2} x[/tex]!

And, technically, you should say [itex]lim_{x\rightarrow 0^+}[/itex] and [itex]lim_{x\rightarrow 0^-}[/itex] but that isn't as bad as the howler above!
 

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