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How can you determine the following limit

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data

    How can you determine the following limit using rationalization?
    1d59dee5b47bea3364b542becd79d9f1.png
    where x = h
    2. Relevant equations


    3. The attempt at a solution
    I attempted to multiply by the conjugate and cannot get the problem to work out. When doing this I noticed no terms canceled out and now I am stuck.

    We have not use l'hopital's rule in our class and I am wondering how I can solve this problem without derivatives.
    I simplified the limit to the following state:

    6cc75e0ad164028c6be328f2189cc694.png
    Additionally, I tried taking the conjugate of the numerator as well. The problem is every substitution leads to 0/0
     
    Last edited: Oct 23, 2015
  2. jcsd
  3. Oct 23, 2015 #2
    Do you mean the following?
    [tex] \lim _{h \to 0} \frac{\sqrt{h^2 + 15} - \sqrt{h + 15}}{\sqrt{h + 3} - 2}[/tex]
     
  4. Oct 23, 2015 #3
    yes
     
  5. Oct 23, 2015 #4

    SteamKing

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    Is the limit supposed to be evaluated as x → 0 or as h → 0?

    In the expression above, there is no x variable.
     
  6. Oct 23, 2015 #5
    Sorry that is a extremely simple mistake you are correct as h--->0
     
  7. Oct 23, 2015 #6
    Have you tried simply plugging [itex]h = 0[/itex] in?
     
  8. Oct 23, 2015 #7
    My bad the problem is actually h----> 1 I have been looking at the problem for 2 hours :p
     
  9. Oct 23, 2015 #8
    You don't necessarily need terms to cancel out when you multiply by the conjugate. What you need to do is ensure that you're not dividing by 0. It would help to show your work when you do that.
     
  10. Oct 23, 2015 #9
    OK I will post a screen shot of my work
     
  11. Oct 23, 2015 #10
     

    Attached Files:

  12. Oct 23, 2015 #11
    So, notice what's causing difficulties. You want to make sure you don't divide by 0, but multiplying by the conjugate of the numerator won't take away that term in the denominator that goes to 0. So perhaps try multiplying by the conjugate of the bottom.
     
  13. Oct 23, 2015 #12

    SteamKing

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    Sometimes, a limit can be evaluated by simply plugging in the limit value, as suggested by axmls. It's only when you get either an infinite or otherwise undefined result by substitution that you need to resort to more subtle manipulations.
     
  14. Oct 23, 2015 #13
    Hi, I tried taking the conjugate of the bottom and a similar problem has occurred:

    BVAxuat.jpg
    The problem is still of the form (0/0) when substituted in
     
  15. Oct 23, 2015 #14
    How can you not divide by zero if the zero term is not eliminated?
     
  16. Oct 23, 2015 #15

    SteamKing

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    Well, this problem keeps changing, for some reason.
     
  17. Oct 23, 2015 #16
    Sorry it should have only been that one change to make it so h---> 1 instead of h---> 0
     
  18. Oct 23, 2015 #17

    Ray Vickson

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    No, no, please don't. Screen shots are discouraged; typing it all out is much preferred, because the results are essentially device-independent.
     
  19. Oct 23, 2015 #18
    I simplified the limit to the following state:

    6cc75e0ad164028c6be328f2189cc694.png
     
    Last edited: Oct 23, 2015
  20. Oct 23, 2015 #19
    The following limit should actually simplify to this:

    4e80e6466c6058dd14e059c382ed55fb.png
     
  21. Oct 23, 2015 #20
    I figured out the solution!! One must multiply the equation by both conjugates!

    Thank you for the Guidance.
     
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