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
$\displaystyle lim_{\theta \rightarrow 0} \frac{\theta}{tan(9\theta)} = \frac{0}{0}$
You can use L'Hopital's rule by taking the derivative of the "top" over the derivative of the "bottom".
$\displaystyle lim_{\theta \rightarrow 0} \frac{1}{9sec^2(\theta)} = \frac{cos^2(9\theta)}{9} \Rightarrow \frac{1}{9}$
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:
$\displaystyle \mathop {\lim }\limits_{\theta \to 0} \frac{\theta }
{{\tan 9\theta }} = \frac{1}
{9}\mathop {\lim }\limits_{\theta \to 0} \frac{{9\theta }}
{{\tan 9\theta }} = \frac{1}
{9}.$