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Math Help - Limits

  1. #1
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    Limits

    how would I find this one, i tried just using smaller and smaller values for theta, but i didnt seem to get the right answer...

    lim theta-->0 (theta)/(tan(9(theta)))

    thanks
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by mathlete View Post
    how would I find this one, i tried just using smaller and smaller values for theta, but i didnt seem to get the right answer...

    lim theta-->0 (theta)/(tan(9(theta)))

    thanks
    lim_{\theta \rightarrow 0} \frac{\theta}{tan(9\theta)} = \frac{0}{0}

    You can use L'Hopital's rule by taking the derivative of the "top" over the derivative of the "bottom".

    lim_{\theta \rightarrow 0} \frac{1}{9sec^2(\theta)} = \frac{cos^2(9\theta)}{9} \Rightarrow \frac{1}{9}
    Last edited by colby2152; January 28th 2008 at 12:41 PM. Reason: typo
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by mathlete View Post
    lim theta-->0 (theta)/(tan(9(theta)))
    You're expected to know that \mathop {\lim }\limits_{\theta  \to 0} \frac{{\tan \theta }}<br />
{\theta } = \mathop {\lim }\limits_{\theta  \to 0} \frac{\theta }<br />
{{\tan \theta }} = 1.

    You need to turn your limit into such form:

    \mathop {\lim }\limits_{\theta  \to 0} \frac{\theta }<br />
{{\tan 9\theta }} = \frac{1}<br />
{9}\mathop {\lim }\limits_{\theta  \to 0} \frac{{9\theta }}<br />
{{\tan 9\theta }} = \frac{1}<br />
{9}.
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