## anonymous 4 years ago lim┬(x→π/2)⁡〖(π(π-2x) tan⁡〖(x-π/2)〗)/(2(x-π) 〖cos〗^2 x)〗

1. anonymous

this is limit of trygonometry, i put it on the ms.word

2. anonymous

Thats indeterminate (0/0) so you would do l'hospital's rule. Have you done that?

3. anonymous

yes, i have done. but, i'm stuck in trygonometry. can you do that?

4. anonymous

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.

5. anonymous

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.

6. anonymous

Thank you

7. anonymous

No problem, sorry its so gross looking :/

8. anonymous

it's okay, you help me anyway :)