1) lim as theata approaches 0 --> 1-cos theata/theata^2
2) lim as theaata approaches 0 --> tan theata/2 theata
Please explain how each is done please. THank yoU! I am supposed to determine the limits for each.
first of all, it's "theta"
it's a lot better than some people though
i don't mind mispelling as much as some people though, but now you know, so you can spell it right next time
one way that everyone will hate is to apply L'Hopital's rule:
thus, $\displaystyle \lim_{\theta \to 0} \frac {1 - \cos \theta}{\theta ^2} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta}$
i think you can take it from there. Hint: think "special limit"
if you want to do it without L'Hopital's, power series is another method. too much work here though, i think
(do you know what L'Hopital's rule is? do you know how to do it using power series?)
$\displaystyle \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta}$ is also a special limt (without the 2 in the denominator that is, but it doesn't matter, we can factor constants out).2) lim as theaata approaches 0 --> tan theata/2 theata
if you want to do it the hard way, you can note that: $\displaystyle \lim_{\theta \to 0} \frac {\tan \theta}{2 \theta} = \lim_{\theta \to 0} \frac {\sin \theta}{2 \theta \cos \theta}$
Hello, superweirdo!
We need the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}\;=\;1$
Multiply by $\displaystyle \frac{1+\cos\theta}{1+\cos\theta}$: . $\displaystyle \frac{1-\cos\theta}{\theta^2}\cdot\frac{1+\cos\theta}{1+\c os\theta} \;=\;\frac{1-\cos^2\!\theta}{\theta^2(1+\cos\theta)} \;=\;\frac{\sin^2\!\theta}{\theta^2(1+\cos\theta)}$$\displaystyle 1)\;\;\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2}$
then: . $\displaystyle \lim_{\theta\to0}\left[\left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac {1}{1+\cos\theta}\right] \;=\;1^2\cdot\frac{1}{1+1} \;=\;\frac{1}{2}$
$\displaystyle 2)\;\;\lim_{\theta\to0}\frac{\tan\theta}{2\theta}$
We have: .$\displaystyle \frac{\frac{\sin\theta}{\cos\theta}}{2\theta}\;=\; \frac{\sin\theta}{2\theta\cos\theta}
$
Then: . $\displaystyle \lim_{\theta\to0}\left[\frac{1}{2}\cdot\frac{\sin\theta}{\theta}\cdot\fra c{1}{\cos\theta}\right] \;=\;\frac{1}{2}\cdot1\cdot\frac{1}{1}\;=\;\frac{1 }{2}$