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Homework Help: Lim x-> 00 sin(1/x)x^2/(2x-1)?

  1. Oct 3, 2008 #1
    lim x-> 00 sin(1/x)x^2/(2x-1)???

    lim x-> 00 sin(1/x)x^2/(2x-1)???

    Iv tried sandwich theorem, l'hopitals and i seem to be going nowhere
     
  2. jcsd
  3. Oct 3, 2008 #2
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Maybe you could use the small angle approximation for sin? (sin(y) = y for small y)
     
  4. Oct 3, 2008 #3
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    But that just makes sin(1/x)=0 which doesnt help, when i graph it i get the limit 1/2 i just dont know how to find it using limit laws
     
  5. Oct 3, 2008 #4
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    You're not applying the small angle approximation. sin(y)=y for small y. I didn't say sin(y)=0 for small y.
     
  6. Oct 3, 2008 #5
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Oh i see that works, but im not too sure if i am meant to use the small angle approximation, but otherwise thanks
     
  7. Oct 3, 2008 #6
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Really? I mean it's just an approximation of the taylor series for sin(x). Seems fair game for a calculus class. Sorry I can't be of more help.
     
  8. Oct 3, 2008 #7

    Gib Z

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    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Personally I don't enjoy justifying that taking sin y = y was accurate enough. It was obvious for cos(e) to question why we were not letting sin y = 0 - You have to take the correct number of terms of the series, and taking more terms until you get the expected answer really isn't good enough imo.

    I would have used the Sandwich theorem with [tex] x - \frac{x^3}{6} \leq \sin x \leq x[/tex].
     
  9. Oct 3, 2008 #8

    HallsofIvy

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    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Since we know that
    [tex]\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex]
    I would be inclined, as a first step, to let y= 1/x. Then the limit becomes
    [tex]\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}[/tex]
    which, multiplying one of the y's from y2 into the parentheses, gives
    [tex]\lim_{y\rightarrow 0}\frac{sin y}{y}\frac{1}{2-y}[/itex]
    which should be pretty easy.
     
  10. Oct 3, 2008 #9

    statdad

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    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Just a question: does "x -> 00" mean the limit as x goes to zero or as x goes to infinity?
     
  11. Oct 3, 2008 #10

    Gib Z

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    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    A very elegant solution Halls =]

    It means the limit as x goes to infinity, commonly used notation on forums for those who don't use latex.
     
  12. Feb 3, 2011 #11
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    derivation to funkton
    f(x)= e^sin^22x when x=pi/2
    i try but i get "0"... is this right solution?
    plz if any one help me
     
  13. Feb 3, 2011 #12

    Mark44

    Staff: Mentor

    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    Please start a new thread rather than tacking your question onto a thread that is more than two years old.
     
  14. Feb 3, 2011 #13
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    com an man im not so used by sympols in com puter .. but i well try
     
  15. Feb 3, 2011 #14
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    calculate the derivative of the function. specify the exact answer

    f(x)= e^sin^2(2x) when x=∏/2
     
  16. Feb 3, 2011 #15

    Mark44

    Staff: Mentor

    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

    I wasn't talking about the symbols - don't use this thread for an unrelated problem. Please start a new thread.
     
  17. Feb 3, 2011 #16
    Re: lim x-> 00 sin(1/x)x^2/(2x-1)???

     
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