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anonymous
 4 years ago
lim x→0 cos 9x − cos^2 *9x / 9x
anonymous
 4 years ago
lim x→0 cos 9x − cos^2 *9x / 9x

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[lim~x\rightarrow \infty\{ \cos(9x) \frac{\cos^2(9x)}{9x}~\text{?}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0on the homework its setup like lim x > 0 ........ cos(9x)  cos^2 9x / 9x

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0break it into smaller parts... 1. \[\lim_{x \rightarrow \infty} \cos(9x)  \lim_{x \rightarrow \infty} \cos^2(9x)/9x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, well you can separate this into two distinct limits, one is very easy to deal with because we know what it turns out to be: like this \[lim~ x \rightarrow 0~~cos(9x)  lim~x \rightarrow 0~~\frac{cos^2(9x)}{9x)}\] the other one resembles the sin function you asked about a few questions ago, the same methods should work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0atleast I think it was you that asked about it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so should it be 1/9 0r 1/81

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0the cos^2(9x) may use l'hopitals rule...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im thinking the 9's match like cos(9x)/ 9x = 1 and then cos 9x *cos 9x / 9x = 1 am i on the right track?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0campbell, no you can't you don't have a case of 0/0,0/infinity,infinity/0 or infinity/infinity, you have 1/0

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0well lim of cos(9x) = 1 as x ==> 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you try plotting this one you can see pretty clearly that as x > 0 the function approaches infinity. there's another pretty useful property of limits that says the limit of a product of two functions is the product of the limits. so what you'd have in this case is \[1\{lim~x\rightarrow 0~(\cos^2(x))\}*\{lim~x\rightarrow0(\frac{1}{9x})\}\] the limit of one of these is one, while the other is infinity, and 1 times infinity is infinity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=limit+as+x+%3E0+of+cos^2%28x%29%2Fx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont think its infinity, i put it into the thing at it said an x, but i think its 3

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0use the idea that that the sum of the limits \[\cos ^{2}(9x) = 1  \sin^2(9x)\] then is \[\lim_{x \rightarrow \infty} \cos(9x)  \lim_{x \rightarrow \infty} (1  \sin^2(9x))/9x\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} \cos(9x)  \lim_{x \rightarrow \infty} 1/9x + \lim_{x \rightarrow \infty} \sin^2(9x)/9x\]
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