1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    No problem :smile:

    This tactic is often used to get rid off square roots.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with a limit
  1. Limits help (Replies: 4)

  2. Help with a limit (Replies: 3)

  3. Limits help! (Replies: 3)

  4. Limits help! (Replies: 3)

  5. Limit help (Replies: 2)

Loading...