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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- Jan 28th 2008, 09:59 AMmathleteLimits
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 - Jan 28th 2008, 10:16 AMcolby2152
$\displaystyle lim_{\theta \rightarrow 0} \frac{\theta}{tan(9\theta)} = \frac{0}{0}$

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

$\displaystyle lim_{\theta \rightarrow 0} \frac{1}{9sec^2(\theta)} = \frac{cos^2(9\theta)}{9} \Rightarrow \frac{1}{9}$ - Jan 28th 2008, 11:05 AMKrizalid
You're expected to know that $\displaystyle \mathop {\lim }\limits_{\theta \to 0} \frac{{\tan \theta }}

{\theta } = \mathop {\lim }\limits_{\theta \to 0} \frac{\theta }

{{\tan \theta }} = 1.$

You need to turn your limit into such form:

$\displaystyle \mathop {\lim }\limits_{\theta \to 0} \frac{\theta }

{{\tan 9\theta }} = \frac{1}

{9}\mathop {\lim }\limits_{\theta \to 0} \frac{{9\theta }}

{{\tan 9\theta }} = \frac{1}

{9}.$