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: Limits

  1. Mar 28, 2007 #1
    URGENT: Limits

    1. The problem statement, all variables and given/known data
    PART A
    Let f(x) be a function defined on an open interval containing c, and let L be a real number. Given:

    [tex]\mathop {\lim }\limits_{x \to a} f\left( x \right) = L[/tex]

    For each real [tex]\varepsilon > 0[/tex] , there exists a real [tex]\delta > 0[/tex] such that for all x:
    [tex]0 < \left| {x - c} \right| < \delta
    0 < \left| {f\left( x \right) - L} \right| < \varepsilon [/tex]

    PART B
    Hence, evaluate exactly [tex]\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta }[/tex]


    2. The attempt at a solution
    I do not have the slightest idea on how to solve part A, however for part B my working is as follows (I am using L'hopital's Rule):
    [tex]
    \begin{array}{l}
    \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \frac{{\sin ^2 4\left( 0 \right)}}{0} = \frac{0}{0} \\
    \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin ^2 4\theta }}{\theta } = \mathop {\lim }\limits_{\theta \to 0} \frac{{f\left( \theta \right)}}{{g\left( \theta \right)}}\quad \Rightarrow \quad f\left( \theta \right) = \sin ^2 4\theta \quad g'\left( \theta \right) = 1 \\
    f\left( \theta \right) = \sin ^2 4\theta \\
    \sin ^2 \theta = \frac{{1 - \cos 2\theta }}{2} \\
    f\left( \theta \right)\sin ^2 4\theta = \frac{{1 - \cos 8\theta }}{2} \\
    \end{array}
    [/tex]

    Am i doing this part correctly? Is there an easier way? Many thanks to all help and suggestions offered.
    unique_pavadrin
     
    Last edited: Mar 28, 2007
  2. jcsd
  3. Mar 28, 2007 #2

    Gib Z

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    Yes you are doing it correctly, but instead of converted the sin squared using double angle formulae, just differentiate as it is. I think that should be easier.
     
  4. Mar 28, 2007 #3
    As far as i can see part A has no question to solve, it simply states the definition of the limit
     
  5. Mar 28, 2007 #4

    Gib Z

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    I think it is asking to prove that statement it provided or something
     
  6. Mar 28, 2007 #5
    Is it not possible to use small angles [tex] {\sin \theta }\approx\theta[/tex]

    so [tex] \frac{{\sin ^2 4\theta }}{\theta }\approx16\theta[/tex] as [tex]\theta \rightarrow 0[/tex]
     
  7. Mar 28, 2007 #6

    Gib Z

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    Well my graphing program says the limit is zero..
     
  8. Mar 28, 2007 #7
    sin^2 4@ == (sin 4@)^2

    then if u do like normal differentiation, u get 8sin4@cos4@
    and with the trigo identity, u will get 4 sin8@
    thus, that result would approaches 0 for @ --> 0

    ps: sorry, dont know how to use the mathematical notation thingy
     
    Last edited: Mar 28, 2007
  9. Mar 28, 2007 #8
    yeah this limit is zero. there is another way, shorter one and simpler one i guess than chickens suggested, although you get the same result. try to multiply both the numerator and denominator by (16@)/(16@), what do you get? then write the numerator as (4*@)^2, so you will get (sin(4@)/4@)^2, and the limit of this is 1, but we have still the limit of 16@ as @->0 so the result is 0.
    sorry for my symbols, i hope u understand @ stands for theta, the angle at sin
     
  10. Mar 28, 2007 #9

    cristo

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    Can you explain this further, as I don't really understand what you mean! Are you doing this? [tex]\lim_{x\rightarrow 0} \frac{\sin^24\theta}{\theta}=\lim \frac{16\theta^2}{\theta}\left(\frac{\sin 4\theta}{4\theta}\right)^2[/tex].

    If so, how do you deduce that [tex]\lim \left(\frac{\sin 4\theta}{4\theta}\right)^2=1[/tex] ?
     
  11. Mar 28, 2007 #10

    HallsofIvy

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    The fact that [tex]\lim_{\theta\rightarrow 0}\frac{sin \theta}{\theta}= 1[/tex] together with the fact that x2 is continuous should do it!
     
  12. Mar 28, 2007 #11

    cristo

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    Indeed. I knew I was missing something pretty obvious!
     
  13. Mar 28, 2007 #12
    thank you for your replies, you have been of great help
    unique_pavadrin
     
  14. Mar 29, 2007 #13
    [tex]\lim_{\theta\rightarrow 0} \frac{\sin^24\theta}{\theta}=\lim_{\theta\rightarrow 0} \frac{16\theta^2}{\theta}=\lim_{\theta\rightarrow 0}16\theta=0[/tex]
     
    Last edited: Mar 29, 2007
  15. Mar 29, 2007 #14

    VietDao29

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    How did you go from the first expression to the second expression? :bugeye:
    [tex]\sin \theta \neq \theta[/tex]
     
  16. Mar 29, 2007 #15
    you need to read the whole threads on this topic explaining how to find this limit. I also explained this on my thread, so read it first, if you still have any questions than after that ask them.
     
  17. Mar 29, 2007 #16

    HallsofIvy

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    The Taylor's expansion for [itex]sin(\theta)[/itex] is
    [tex]\theta- \frac{1}{2}\theta^2+ \frac{1}{3!}\theta^3- \cdot\cdot\cdot[/itex]

    The "linear approximation" (tangent line) for [itex]sin(\theta)[/itex] is just [itex]\theta[/itex] so for [itex]sin(4\theta)[/itex] is [itex]4\theta[/itex] and for [itex]sin^2(4\theta)[/itex] is [itex]16\theta^2[/itex].

    Since the "higher order" terms will go to 0 faster than the first order term, it is sufficient to use the linear approximate to find the limit. (It is really the same as using L'Hopital's rule.)
     
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