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Unsolvable limit of trigonometric function

  1. Dec 8, 2008 #1
    I need to find the limit of sin(3t)^2/t^2 as t approaches 0. We have not yet learned L'Hopital's rule, so how do I find the limit here?

    I tried to take the derivative of sin(3t)^2/t^2, but it is nowhere near cancelling out 't' from the bottom.

  2. jcsd
  3. Dec 8, 2008 #2


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    Do you know the limit of sin(x)/x as x->0? Do you know the rules for a limit of a product?

    If, so use those along with the fact that [tex]\frac{\sin^2(3t)}{t^2}=\frac{\sin(3t)}{t}\cdot\frac{\sin(3t)}{t}[/tex]
  4. Dec 9, 2008 #3


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    And, if that isn't enough, the obvious
    [tex]\frac{sin^2(3t)}{t^2}= 9\frac{sin(3t)}{3t}\cdot\frac{sin(3t)}{3t}[/tex]
  5. Dec 9, 2008 #4
    [tex]\lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})[/tex]

    Now remember [tex]\lim_{t \rightarrow 0}\frac{sin(t)}{t}=1[/tex].

    So [tex]\lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})=\lim_{t \rightarrow 0}(\frac{sin(3t)}{t})^2[/tex].

    It is pretty easy from now on. In future, please consider using LaTeX code.

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