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dinainjune
 4 years ago
lim┬(x→π/2)〖(π(π2x) tan〖(xπ/2)〗)/(2(xπ) 〖cos〗^2 x)〗
dinainjune
 4 years ago
lim┬(x→π/2)〖(π(π2x) tan〖(xπ/2)〗)/(2(xπ) 〖cos〗^2 x)〗

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dinainjune
 4 years ago
Best ResponseYou've already chosen the best response.0this is limit of trygonometry, i put it on the ms.word

malevolence19
 4 years ago
Best ResponseYou've already chosen the best response.1Thats indeterminate (0/0) so you would do l'hospital's rule. Have you done that?

dinainjune
 4 years ago
Best ResponseYou've already chosen the best response.0yes, i have done. but, i'm stuck in trygonometry. can you do that?

malevolence19
 4 years ago
Best ResponseYou've already chosen the best response.1The 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: \[\frac{\pi}{2}\lim_{x \rightarrow \frac{\pi}{2}}\frac{(2)(\tan(x\frac{\pi}{2}))+(\pi2x)\sec(x\frac{\pi}{2})\tan(x\frac{\pi}{2})}{\cos^2(x)+(x\pi)(2)(\sin(x))\cos(x))}\] However you notice this is still undeterminate. (0/0) so you have to differentiate it again.

malevolence19
 4 years ago
Best ResponseYou've already chosen the best response.1So differentiate it again: \[\frac{\pi}{2}\lim_{x \rightarrow \frac{\pi}{2}}\frac{2 \pi \csc^3(x) (2 \sin(x)+(\pi2 x) \cos(x))}{4 (\pix) \cos(2 x)4 \sin(2 x)}\] Evaluating this should give you 2.

malevolence19
 4 years ago
Best ResponseYou've already chosen the best response.1No problem, sorry its so gross looking :/

dinainjune
 4 years ago
Best ResponseYou've already chosen the best response.0it's okay, you help me anyway :)
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