Finding the limit of the equation

  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. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    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. 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. 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. Dick

    Dick 25,893
    Science Advisor
    Homework Helper

    Yes. The limits from the two sides are different. So there is no common limit. So the limit doesn't exist.
     
  7. OMG! I can't believe it was that simple... :(
     
  8. 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. Gib Z

    Gib Z 3,348
    Homework Helper

    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. HallsofIvy

    HallsofIvy 40,964
    Staff Emeritus
    Science Advisor

    [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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