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Homework Help: Finding the limit of the equation

  1. Mar 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate the limit if it exists

    2. Relevant equations

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

    3. 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: Mar 4, 2007
  2. jcsd
  3. Mar 4, 2007 #2


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    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.
  4. Mar 4, 2007 #3
    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.
    Last edited: Mar 4, 2007
  5. Mar 4, 2007 #4
    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?
  6. Mar 4, 2007 #5


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    Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.
  7. Mar 4, 2007 #6
    OMG! I can't believe it was that simple... :(
  8. Mar 4, 2007 #7
    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.
  9. Mar 5, 2007 #8

    Gib Z

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    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.
  10. Mar 5, 2007 #9


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    [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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