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Homework Help: Help with a limit

  1. Aug 27, 2005 #1
    Hi,

    Lim(t->0) t / sqrt(4+t)-sqrt(4-t)

    I've stared at this for like half an hour :( Could someone please give me some hints of how I start factorizing this? Thanks.
     
  2. jcsd
  3. Aug 27, 2005 #2
    Can you use L'Hopital's Rule?
     
  4. Aug 27, 2005 #3
    No, sorry. We havnent gotten that far yet :)
     
  5. Aug 27, 2005 #4

    TD

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    Multiply nominator & denominator with the complement of the denominator, i.e. [itex]\left(\sqrt {4 + t} + \sqrt {4 - t}\right)[/itex]

    In the denominator, use [itex]\left( {a - b} \right)\left( {a + b} \right) = a^2 - b^2[/itex]

    Now, you should be able to cancel out a t and just fill in t = 0.
     
  6. Aug 27, 2005 #5

    Curious3141

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    The easiest way, use the binomial expansion (the general form for nonintegral exponents).

    [tex](1 + x)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x[/tex] for [itex]|x| << 1[/itex]

    So

    [tex]\lim_{t -> 0}\frac{t}{\sqrt{4 + t} - \sqrt{4 - t}} = \lim_{t -> 0}\frac{t}{2(\sqrt{1 + \frac{t}{4}} - \sqrt{1 - \frac{t}{4}})} = \lim_{t -> 0}\frac{t}{2[(1 + \frac{t}{8}) - (1 - \frac{t}{8})]} = \lim_{t -> 0}\frac{t}{\frac{t}{2}} = 2[/tex]
     
  7. Aug 27, 2005 #6
    TD: Then i get ( sqrt(4+t)-sqrt(4-t)) / 2

    But the answer is supposed to be 2. I'm still stuck :(
     
  8. Aug 27, 2005 #7

    TD

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    It should be a + between those square roots in the nominator :wink:
     
  9. Aug 27, 2005 #8
    Curious: I've never seen that formula before. But thanks.
     
  10. Aug 27, 2005 #9
    TD: Gaaah, stupid me. Well I got it now. Thanks!
     
  11. Aug 27, 2005 #10

    TD

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    No problem :smile:

    This tactic is often used to get rid off square roots.
     
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