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anonymous
 2 years ago
LIMIT!
using L'Hopital's rules, evaluate
lim x appraoches infinity (cos(x/2))^x^2
anonymous
 2 years ago
LIMIT! using L'Hopital's rules, evaluate lim x appraoches infinity (cos(x/2))^x^2

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myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Do you know you can write y as e^ln(y) let me know if that helps brb

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1so what I'm say is you can write \[(\cos(x/2))^{x^2}=e^{\ln((\cos(x/2))^{x^2})}=e^{x^2 \ln(\cos(x/2))}=e^{\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i think i forgot about it

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1now you did to find this limit \[\lim_{x \rightarrow \infty}\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wow, seems it is so complicated.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so, we can just take the value of power of exponential? what say you?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Instead of writing \[x^2 \ln(\cos(x/2)) \text{ as } \frac{\ln(\cos(x/2))}{\frac{1}{x^2}} \text{ it might be more preferable } \\ \text{ if we write is as } \frac{x^2}{\frac{1}{\ln(\cos(x/2))}}\] still don't know if we should use l'hospital even though it says and that is why I attempted to set the problem up in this way the reason I say that is because the limit as x approaches inf of ln(cos(x/2)) dne since lim x approaches inf of cos(x/2) does not exist since it will oscillate between 1 and 1 but pretending we can use l'hospital since it say use l'hospital i guess we will force the l'hospital

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1@zepdrix the reason i ask is because i only see it used for cases like inf/inf or 0/0

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Oh boy D: Good question.. hmm So we have a problem since ln(cosx) is undefined over and over as x>infinity, yah?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Grr I dunno >.< maybe lego man knows

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1maybe sith knows i seen him around a lot doing math like it is nothing

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Wolfram seems to agree; the limit doesn't exist. That's my guess

sweetburger
 2 years ago
Best ResponseYou've already chosen the best response.0LHops international house of limits

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Not only does wolfram not say anything, but Mathematica comes up blank as well.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Does that mean Mathematica doesn't have the programming to find the limit or that the limit does not exist?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1i believe the limit does not exist and i also believe we don't need l'hopital (even though it says to use it)

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1I think it only tells you that mathematica doesn't have the programming.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm... power series? With WA, adding more terms to the cosine series makes the limit alternate between infinity and complex infinity... Not sure what that tells us.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Can it converge if it is only intermittently continuous? This is not some countably infinite number of discontinuities (where some Lebesgue measure might provide consistent results). In the logarithm form, there are giant gaps. This is really why introducing the logarithm is a bad idea. It massively modifies the Domain. If you can tell me there is a big number, M, where \(cos(x/2) > 0\;for\;x > M\), then the logarithm is of no concern. There is no such number M.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1This limit can't possibly exist. Since cosine regularly alternates between \(\pm1\), no matter what value of \(x\) we're currently looking at, we can get a larger value such that \(\cos(x/2)\) is 1. Powers have no effect on this, and so you can always get a larger value such that the function outputs 1. But you can also get a larger value whose output will give you 0. So there can't possibly be a limit.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Sequence: \((\cos(4n\pi))^{n^2}\;for\; n\in\mathbb{N}\) It's just ones (1s).

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I believe the question is not properly phrase. As x tends infinity, cos(x/2) and sin(x/2) is not defined.
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