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anonymous
 one year ago
I don't believe that I solved this problem correctly...
(will post below)
anonymous
 one year ago
I don't believe that I solved this problem correctly... (will post below)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \cos2\theta }{ \cos \theta \sin \theta}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know that \[\cos2 \theta = 12\sin^2\] so I changed that to the numerator, and then plugged in my limit into the thetas, which is pi/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I ended up with 0/0 meaning that the limit didn't exist, but I have a feeling I did this wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try the identity\[\cos 2\theta = \cos^2\theta  \sin^2\theta\]This can be factored and then the fraction can be simplified

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me clarify... pi/4 = c not the limit. Sorry about that.

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\cos ^2\theta\sin ^2\theta}{\cos \theta\sin \theta}=\frac{(\cos \theta\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta\sin \theta}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I would just cancel out the like terms in the numerator and the denominator, plug in my c values and solve?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, thank you. :) I guess I'll medal ospreytriple since they responded first? But I'll fan both of you. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not to butt in but if you get \[\frac{0}{0}\] it does NOT mean the limit doesn't exist it means you have to do more work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great to know! I'll keep that in mind next time I get it as an answer.
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