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

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

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Other answers:

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.
Thank you
No problem, sorry its so gross looking :/
it's okay, you help me anyway :)

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