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Homework Help: Something simple that i cannot visualize?

  1. Sep 11, 2007 #1
    Something simple that i cannot visualize!?!?

    1. The problem statement, all variables and given/known data

    limit as x approaches 0 for sinx/4x

    2. Relevant equations

    ok....what i know is that sinx/x=1 so by that simple statement the answer becomes 1/4 because sinx/x=1 and than there's 4x which gives me the 4....so 1/4......i tend to over think things......another question is why would cos3x=1 when x goes to 0? i know cosx goes to 1 when x equals 0.....but wouldn't you multiply 1 by 3 so it would be 3 or is it because it oscillates to 1 all the time?


    3. The attempt at a solution

    ok so there's a way our teacher does it that i'm trying to figure out that way, even though i know the answer right off the bat, just thinking it might help me in the future

    lim as X approaches 0 for sinx/4x.........lim as x approaches 0 for (sinx)/((4x/4)*4)....this is to get rid of the 4(or should i be multiplying by 4x?) in the 4x and to put it on the outside.......1/4 lim as x approaches 0 for (sinx)/(x).......lim as x approaches 0 for (1/4)*(1)=1/4.....look right to everyone? am i cancelling out the 4x right?
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2
  4. Sep 12, 2007 #3
    Hint: ignore squeeze theorem, use l'Hôpital's rule
     
    Last edited: Sep 12, 2007
  5. Sep 12, 2007 #4

    CompuChip

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    Homework Helper

    By L'Hopital,
    [tex]\lim_{x \to 0} \frac{\sin x}{4 x} = \lim_{x \to 0} \frac{\cos x}{4} = \frac{1}{4}[/tex]

    Or by Taylor expansion,
    [tex]\lim_{x \to 0} \frac{\sin x}{4 x} = \lim_{x \to 0} \frac{x + \mathcal O(x^3) }{4 x} = \lim_{x \to 0} \frac{1 + \mathcal O(x^2) }{4} = \lim_{x \to 0} \left( \frac{1}{4} + \mathcal O(x^2) \right) = \frac14.[/tex]

    Both ways completely analogous to how you would prove the [itex]\sin x/x[/itex] limit.

    Or the stupid way :smile:
    [tex]\lim_{x \to 0} \frac14 \frac{\sin x}{x} = \left( \lim_{x \to 0} \frac14 \right) \times \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac14 \times 1 = \frac14. [/tex]
     
  6. Sep 12, 2007 #5
    :rofl:
     
    Last edited: Sep 12, 2007
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