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

1. dinainjune

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

2. malevolence19

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

3. dinainjune

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

4. malevolence19

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. malevolence19

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. dinainjune

Thank you

7. malevolence19

No problem, sorry its so gross looking :/

8. dinainjune

it's okay, you help me anyway :)