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Simple Limits Question

  1. Oct 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following limit
    [itex]
    \displaystyle\lim_{x\rightarrow 4} {\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}}[/itex]
    2. Relevant equations

    lim x -> 4
    3. The attempt at a solution
    i tried change of variables so i get [itex]√2x+1 = u[/itex]
    then i rearranged that to get [itex]u^2-1/2[/itex]
    And i rearrnage original equation and i get this..
    [itex]-(u^2-7)/3-u(u^2-1/2)[/itex]

    Sorry i am new with this latex thing .. i don treally konw how to use it the sqrt thing so i just lay it out this way
     
    Last edited by a moderator: Oct 6, 2014
  2. jcsd
  3. Oct 6, 2014 #2
    I am really unsure of how to use this latex thing... but personally i think my attempt at the solution is all garbage so if anyone could guide me on the right path that would be help ful
     
  4. Oct 6, 2014 #3
    Multiply with the nominator and denominator with 3+sqrt(4x+1)
     
  5. Oct 6, 2014 #4

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Or l'hopital's?
    ##\frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}\frac{3+\sqrt{2x+1}}{3+\sqrt{2x+1}}=\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{9 - (2x +1)}##
    which still has a divide by zero error as x->4
     
  6. Oct 6, 2014 #5
    i think l'hopitals is derivatives, i havent learned that yet im only allowed to use limits
     
  7. Oct 6, 2014 #6

    Mark44

    Staff: Mentor

    You need to rationalize both the numerator and denominator. To do this, multiply by the following:
    $$ \frac{2 + \sqrt{x}}{2 + \sqrt{x}} \cdot \frac{3 + \sqrt{2x + 1}}{3 + \sqrt{2x + 1}}$$

    When you do this, you get something that you can simplify and then take the limit.
     
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