Here's the question you clicked on:
dinainjune
lim┬(x→π/2)〖(π(π-2x) tan〖(x-π/2)〗)/(2(x-π) 〖cos〗^2 x)〗
this is limit of trygonometry, i put it on the ms.word
Thats indeterminate (0/0) so you would do l'hospital's rule. Have you done that?
yes, i have done. but, i'm stuck in trygonometry. can you do that?
The functions involved in the limit are arbitrary as long as they are easily differentiable and the limits actually exist. So taking the derivatives you have: \[\frac{\pi}{2}\lim_{x \rightarrow \frac{\pi}{2}}\frac{(-2)(\tan(x-\frac{\pi}{2}))+(\pi-2x)\sec(x-\frac{\pi}{2})\tan(x-\frac{\pi}{2})}{\cos^2(x)+(x-\pi)(2)(-\sin(x))\cos(x))}\] However you notice this is still undeterminate. (0/0) so you have to differentiate it again.
So differentiate it again: \[\frac{\pi}{2}\lim_{x \rightarrow \frac{\pi}{2}}\frac{-2 \pi \csc^3(x) (2 \sin(x)+(\pi-2 x) \cos(x))}{4 (\pi-x) \cos(2 x)-4 \sin(2 x)}\] Evaluating this should give you 2.
No problem, sorry its so gross looking :/
it's okay, you help me anyway :)