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Homework Help: Finding lim of following question

  1. Jan 4, 2005 #1
    I am having trouble finding the lim of the following question. Any help?

    lim
    theta -> 0+

    (theta - sin theta) / 2 ( tan (1/2)theta - 1/2 theta)

    The answer is suppose to be 2, but I keep on getting 1.
    Thanks in advanc,e
     
  2. jcsd
  3. Jan 4, 2005 #2

    Hurkyl

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    What have you done?
     
  4. Jan 4, 2005 #3
    I have done the following, but don't know where to go afterwards.

    1 - cos theta / sec^2 (1/2) theta
    = 1 - cos theta / 1 + tan^2 (1/2) theta

    However, it's not in the form 0/0, so i can't apply L'Hopital anymore. Any idea?
     
  5. Jan 4, 2005 #4
    you mean [tex]\frac{\theta - sin(\theta)}{2(tan(\frac{\theta}{2}) - \frac{\theta}{2})}[/tex]

    If so, just apply de l'Hôpital two times in a row and you case is done...


    regards
    marlon
     
  6. Jan 4, 2005 #5

    Zurtex

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    It may not be in the form 0/0 but can't you let theta=0 anyway? Also in the line above couldn't you apply the fact that sec theta = 1 / cos theta?

    I don't know I've never really had to take the limits of anything.
     
  7. Jan 4, 2005 #6

    dextercioby

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    Applying 2 times the rule of Antoine L'Ho^spital very,very carefully,u get '2'.Trust me.

    An elegant alternative proof (i like it more than L'Ho^spital):
    [tex] \lim_{\theta\searrow 0} \frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})} [/tex](1)

    Since the limits is to 0,we know for about 300 years that,close to zero
    [tex] \sin\theta\sim \theta+\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!} [/tex](2)
    [tex] \cos\theta\sim 1+\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!} [/tex](3)

    From (2) and (3) i use only the first 2 terms.I plug these expressions into (1),after expressing '\tan' as a ratio between 'sin' and 'cos'.
    [tex]\lim_{\theta\searrow 0} \frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})}=\lim_{\theta\searrow 0}\frac{\theta-\theta-\frac{\theta^{3}}{3!}}{2[\frac{\theta^{3}}{8\cdot 3!}-\frac{\theta}{2}(1+\frac{\theta^{2}}{4\cdot 2!})]}\cos\frac{\theta}{2} [/tex]
    [tex]=\lim_{\theta\searrow 0}\frac{-\frac{\theta^{3}}{3!}}{2(\frac{\theta^{3}}{8\cdot 3!}-\frac{\theta^{3}}{8\cdot 2!})}\cos\frac{\theta}{2} [/tex](4)


    Simplify through the function at the numerator and find
    [tex] \lim_{\theta\searrow 0}\frac{\theta-\sin\theta}{2(\tan\frac{\theta}{2}-\frac{\theta}{2})}=\lim_{\theta\searrow 0}\frac{1}{2(-\frac{1}{8}+\frac{3}{8})}\cos\frac{\theta}{2}=\frac{8}{4}=2 [/tex](5)

    Daniel.

    PS.Math is beauty... :approve:
     
  8. Jan 4, 2005 #7

    Hurkyl

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    I assume you mean that you applied L'Hôpital's rule to:

    [tex]
    \frac{\theta - \sin \theta}{2(\tan (1/2)\theta - (1/2)\theta)}
    [/tex]

    and got

    [tex]
    \frac{1 - \cos \theta}{\sec^2 (1/2)\theta}
    [/tex]

    First comment: How did you get 1? This limit goes to 0 as θ approaches 0. (numerator goes to 0, denominator goes to 1)

    Second comment: For some reason, you know your answer is wrong. Well, you've only done three things:

    (1) Decided to apply L'Hôpital's rule
    (2) Differentiated the numerator
    (3) Differentiated the denominator

    Logic dictates that you made a mistake in one of these...


    P.S. please use parentheses around your numerators and denominators. In proper usage, 1 + 2 / 3 + 4 means [itex]1 + \frac{2}{3} + 4[/itex], not [itex]\frac{1 + 2}{3 + 4}[/itex].
     
    Last edited: Jan 4, 2005
  9. Jan 5, 2005 #8
    Got it. Thanks for your help.
     
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