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Homework Help: Limits: Zero / Non-Zero

  1. Oct 16, 2008 #1

    I'm trying to find the following limit:
    [tex]{\lim_{x \to 5}}\ {5-x \over {3-\sqrt{x^2 -16}}[/tex]

    I tried 2 things
    Simplifying the bottom:
    [tex]3-\sqrt{x^2-16}[/tex] = [tex]3-\sqrt{x^2-4^2}[/tex] = 3-x-4 = -1-x

    But that doesn't help with what's on top...

    I also tried multiplying top and bottom by:
    [tex]3+\sqrt{x^2-16}[/tex] but I still ended up with 0/non-zero.

    Could anyone point me into the right direction, or correct me if one of those 2 steps was right?

  2. jcsd
  3. Oct 16, 2008 #2


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    How do you figure Sqrt(x^2 - 4^2) = x - 4?

    Have you thought of differentiating the numerator and the denominator, then taking the limit of their ratio?
  4. Oct 16, 2008 #3
    If the limit is zero over non-zero, the limit is just zero. But in this case I'm getting the limit to be 0/0, yeah?
  5. Oct 17, 2008 #4


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    This is wrong. [itex]\sqrt{x^2- 4^2}[/itex] is NOT x- 4.

    Good! That prevented you from getting the wrong answer!

    Really? WHAT did you get when you did that? I do NOT get "0/non-zero".

    I recommend you recheck your algebra. In particular, what is
    [tex]\left(\frac{5-x}{3-\sqrt{x^2- 16}}\right)\left(\frac{3+\sqrt{x^2- 16}}{3+\sqrt{x^2- 16}}\right)[/tex]

    By the way, as MATdaveLACK said, there is nothing wrong with getting "0/non-zero": that would just mean the limit is 0. However, here, that is wrong.
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