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Limits involving sin.

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    find the limit:

    [tex]\lim_{x\rightarrow\frac{\pi}{6}}\frac{2sin(x)-1}{6x-\pi}[/tex]

    3. The attempt at a solution

    Any starters please, I've tried the substitution method but went nowhere.I did x=(pi+h)/6. I still get 0 over something which is zero and i know that is not correct.
     
    Last edited: Feb 27, 2012
  2. jcsd
  3. Feb 27, 2012 #2

    jbunniii

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    Can you please write out what you got when you tried the substitution?
     
  4. Feb 27, 2012 #3
    i got :

    [tex]\lim_{x\rightarrow 0}\frac{2sin(\frac{\pi}{6}+\frac{h}{6})-1}{h}=\frac{0}{0}[/tex]

    and that's undefined.
     
    Last edited by a moderator: Feb 27, 2012
  5. Feb 27, 2012 #4
    Any help please?
     
  6. Feb 27, 2012 #5

    Mark44

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    This is an easy problem if you can use L'Hopital's Rule. I've tried several other approaches, but I have not hit on anything else that worked. That's not to say that there isn't another approach - I just wasn't able to stumble onto it.
     
  7. Feb 27, 2012 #6
    Yea with l'hopital's rule i come up with cos(x)/3 which then is easy to compute. But I have no clue on how to solve it without that.
     
  8. Feb 27, 2012 #7

    Mark44

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    There probably is a way, but I have not been successful finding it. If you are allowed to use L'Hopital's Rule, I would use it.
     
  9. Feb 27, 2012 #8
    Yes, but too bad i can't use it. You've tried the substitution method and you keep getting 0/0 as well right?
     
  10. Feb 27, 2012 #9

    Dick

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    Try expressing the limit in terms of u=x-pi/6 as u->0. You should know the limits of sin(u)/u and (1-cos(u))/u.
     
  11. Feb 27, 2012 #10
    Yes, but that just keeps giving 0/0 which is undetermined.
     
  12. Feb 27, 2012 #11

    Dick

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    sin(u)/u is 0/0 but the limit as u->0 is 1. Just because it's 0/0 doesn't mean there is no limit.
     
  13. Feb 27, 2012 #12
    Yes i understand that part i know that lim x->0 sin(u)/U=1 but i don't know where i'm going to get the sin and the 1-cos(x) so i do (1-cos(x))/x^2=1/2. I have no clue how to reach that.
     
  14. Feb 27, 2012 #13

    Dick

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    After the substitution sin(x) turns into sin(u+pi/6). Use the sin addition formula to break that up.
     
  15. Feb 27, 2012 #14
    ok i got:

    [tex]\lim_{u\rightarrow0}\frac{2(sin(u)*cos(pi/6)+cos(u)*sin(pi/6))-1}{6u}[/tex]

    and when i sub in with 0 i got 0/0.
     
  16. Feb 27, 2012 #15

    Dick

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    Stop subbing in. Of course, it's still 0/0. What is sin(pi/6)? I do think you can express that in terms of limits of sin(u)/u and (1-cos(u))/u if you try.
     
  17. Feb 27, 2012 #16
    Of course i know sin(pi/6) is 1/2 but i don't get how you want me to express it.
     
  18. Feb 27, 2012 #17

    Dick

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    I would like you to express it as (something)*sin(u)/u+(something else)*(1-cos(u))/u.
     
  19. Feb 27, 2012 #18
    I understand what you mean 100% but I don't know where i'm going to get the 1-cos(x) from.
     
  20. Feb 27, 2012 #19

    Dick

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    One of the terms in the numerator is 2*cos(u)*sin(pi/6), right?
     
  21. Feb 27, 2012 #20

    SammyS

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    What's 2 times (1/2) ?
     
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