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Solving Limits that can't be rationalized (i think)

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \lim_{x\to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right) [/tex]

    [tex]
    \lim_{x\to 0}\left(\frac{2^2^x {-2^x}}{2^x{-1}}\right) [/tex]


    3. The attempt at a solution

    I tried rationalizing but doesn't really help in evaluating these limits.


    If a limit cannot be factored, what other ways are there of evaluating limits?
     
  2. jcsd
  3. Dec 10, 2008 #2

    Mentallic

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    For the first I can't seem to figure it out either. The denominator can be rationalised of course, but what comes after that I'm unsure.

    As for the second, notice that [tex]2^{2x}=(2^x)^2[/tex] since [tex](a^b)^n=a^{bn}[/tex]
    From there you can factorise and simplify :wink:
     
  4. Dec 10, 2008 #3
    Ah, okay, I see how to solve the 2nd one now, thanks =).

    Hopefully someone knows how to solve the 1st one
     
  5. Dec 10, 2008 #4

    gabbagabbahey

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    Have you learned L'Hopital's rule yet? If so, the first limit is fairly simple with that method.
     
  6. Dec 10, 2008 #5
    We have not learned L'Hopital's rule or derivatives yet =(
     
  7. Dec 10, 2008 #6

    gabbagabbahey

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    In that case try multiplying both the numerator and denominator by [itex](2+\sqrt{x})(3+\sqrt{2x+1})[/itex] and then simplify.....this should remove that pesky 0/0.
     
  8. Dec 12, 2008 #7
    note im not thread stater
    when i apply l'Hôpital's i will get 0.75
    however if i dont i get infinity
     
  9. Dec 12, 2008 #8

    gabbagabbahey

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    0.75=3/4 is corrrect. You shouldn't be getting infinity using my method....what is [tex]\left(\frac{(2-\sqrt{x})(2+\sqrt{x})}{(3-\sqrt{2x+1})(3+\sqrt{2x+1})}\right) \left( \frac{3+\sqrt{2x+1}}{2+\sqrt{x}} \right)[/tex]? :wink:
     
  10. Dec 12, 2008 #9
    opps. my bad. miscalculated
     
  11. Dec 12, 2008 #10
    [tex]\lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{2+\sqrt{x}}{3+\sqrt{2x+1}}*\frac{3+\sqrt{2x+1}}{2+\sqrt{x}}[/tex]

    Or:

    [tex]\lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{2+\sqrt{x}}{3+\sqrt{2x+1}}*\frac{3}{2}[/tex]

    Regards.
     
  12. Dec 12, 2008 #11
    Thanks for the help everyone.

    Later, I also found it to work with using this:
    [tex]
    \lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{3+\sqrt{2x+1}}{3+\sqrt{2x+1}}[/tex]
    Simplifies to
    [tex]
    \lim_{x \to 4}\left(\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{8-2x}}\right)[/tex]
    to
    [tex]
    \lim_{x \to 4}\left(\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{2(2-\sqrt{x})(2+\sqrt{x})}}\right)[/tex]

    dividing out the 2-sqrt(x) results in:
    [tex]
    \lim_{x \to 4}\left(\frac{3+\sqrt{2x+1})}{2(2+\sqrt{x})}\right)[/tex]

    after which I just subbed in 4 for the answer 3/4

    edit: does anyone know what this method is called? multiplying by the ______ (what i did to the original equation)
     
    Last edited: Dec 12, 2008
  13. Dec 13, 2008 #12
    You did rationalization of the denominator.

    Regards.
     
  14. Dec 13, 2008 #13

    Mentallic

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    You multiplied by its conjugate.
     
  15. Dec 13, 2008 #14
    Thats the one. Thanks =)
     
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