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Limit :|

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    May seem easy to people but I have no idea how to do this :frown:
    [itex]lim_{x\rightarrow0}\frac{sin x}{2x^{2}-x}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    I factored out an x from the bottom but that really makes me see nothing else to do.. So I tried the quotient rule and that lead me to the wrong answer..
     
  2. jcsd
  3. Sep 15, 2011 #2

    gb7nash

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    Do you know l'hopital's rule?
     
  4. Sep 15, 2011 #3
    Unfortunately no.. I've heard that's what I need for this but we haven't learned that yet..
     
  5. Sep 15, 2011 #4

    gb7nash

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    Well, you don't need it for this problem, but it makes the problem much easier to solve. Did you learn taylor series yet?
     
  6. Sep 15, 2011 #5
    No lol this is Calc 1 basically the 2nd week of class :p
     
  7. Sep 15, 2011 #6

    gb7nash

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    Maybe a better question to ask is, what have you learned with limits?
     
  8. Sep 15, 2011 #7
    Nvm figured it out, if anyone needs this answer PM me.
     
    Last edited: Sep 15, 2011
  9. Sep 15, 2011 #8

    HallsofIvy

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    [tex]\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\left(2x-1\right)[/tex]
    and
    [tex]\lim_{x\to 0}\frac{sin(x)}{x}[/tex]
    is a well-known limit.
     
  10. Sep 15, 2011 #9
    Yep that's what I ended up doing, thanks tho :D
     
  11. Sep 15, 2011 #10

    Mark44

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    That first line above should be

    [tex]\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\frac{1}{2x-1}[/tex]
     
  12. Sep 15, 2011 #11

    symbolipoint

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    At least HallOfIvy showed his work so we could understand the method. Nice doing, both you.
     
  13. Sep 16, 2011 #12
    Or you could have done the squeeze theorem which will be useful in your later on classes:

    [itex]\frac{-1}{2x^{2}-x} \leq \frac{sin(x)}{2x^{2}-x} \leq \frac{1}{2x^{2}-x}[/itex]

    You can easily solve both the left and right side and see that they're both the same number, so the one in the middle has to be that number.
     
  14. Sep 16, 2011 #13

    Mark44

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    This technique DOES NOT apply in this problem, because both ends of the inequality are undefined at x = 0. The squeeze theorem is useful only if the two bounding functions have limits.
     
  15. Sep 16, 2011 #14

    HallsofIvy

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    ??? No, they are NOT the same number!
     
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