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Limit of Absolute Value of X

  1. Sep 21, 2008 #1
    Here, it says that for the limit f(x) = |x| / x,

    |x| = { x, x > 0
    -x, x < 0 }

    What I don't undestand is why is |x| = -x for values under zero? Isn't the absolute value for negative values just x and not -x?


    EDIT: I don't want to start a new thread, but I got stuck on this next question :(

    Lim x approaching -1 of
    3(1-x^2) / x^3 + 1

    I tried multiply the equation by x^3 - 1 / x^3 - 1
    and I ended up with 0 / -2.

    The answer is however 2... any help would be helpful;!

    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2


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    Example: |-7| = -(-7) = 7.

    Do you see why we need the minus sign?
  4. Sep 21, 2008 #3
    Thanks a lot! I can't believe I cannot see such a simple logic...
  5. Sep 21, 2008 #4


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    Ofcourse, what you could do is, by knowing in forhand the limit, to use the delta-epsilon method. But that would become quite messy, I think.

    Your strategy of multiplying the limit with 1 with apropriate numerator and denominator is clever, and is often useful when you deal with square roots.

    Here however, you see that the denominator and numerator have the same limit: 0. That's an indication to use L'hopital. As far as I can see your limit is equal to


    lim_{x \rightarrow -1} \frac{3-3x^{2}}{1+x^{3}} = lim_{x \rightarrow -1}\frac{-6x}{3x^{2}} = lim_{x \rightarrow -1} \frac{-2}{x} = 2
  6. Sep 21, 2008 #5
    Thanks for the help! I was stuck on this equation for a while.
    It's been about a year since I've even touched Calculus... I'm extremely rusty on the basics, and now I have to relearn everything!
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