another trig limit problem (but this time I think I know what im doing)

Hello, kingsolomonsgrave!

Quote:

$\displaystyle\lim_{\theta\to0}\frac{\cos\theta-1}{\sin\theta}$

Can I divide the numerator and denominator by theta and thus get the form 0/1=0 ?

Yes, you can! . . . Good thinking!

Another approach:

$\frac{\cos\theta - 1}{\sin\theta} \;=\;\frac{-(1-\cos\theta)}{\sin\theta} \;=\;-\frac{1-\cos\theta}{\sin\theta}\cdot\frac{1+\cos\theta}{1+ \cos\theta} \;=\;-\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)}$

. . . . . . . $=\;-\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)} \;=\; \frac{-\sin\theta}{1+\cos\theta}$

Therefore: . $\displaystyle \lim_{\theta\to0}\frac{-\sin\theta}{1+\cos\theta} \;=\;\frac{0}{2} \;=\;0$