anonymous
  • anonymous
LIMIT! using L'Hopital's rules, evaluate lim x appraoches infinity (cos(x/2))^x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
Do you know you can write y as e^ln(y) let me know if that helps brb
myininaya
  • myininaya
so 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
  • anonymous
i think i forgot about it

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myininaya
  • myininaya
now you did to find this limit \[\lim_{x \rightarrow \infty}\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}\]
anonymous
  • anonymous
wow, seems it is so complicated.
anonymous
  • anonymous
so, we can just take the value of power of exponential? what say you?
anonymous
  • anonymous
@myininaya
myininaya
  • myininaya
Instead 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
  • myininaya
@zepdrix the reason i ask is because i only see it used for cases like inf/inf or 0/0
myininaya
  • myininaya
not 0/dne or dne/inf
zepdrix
  • zepdrix
Oh boy D: Good question.. hmm So we have a problem since ln(cosx) is undefined over and over as x->infinity, yah?
myininaya
  • myininaya
yah
zepdrix
  • zepdrix
Grr I dunno >.< maybe lego man knows
myininaya
  • myininaya
maybe sith knows i seen him around a lot doing math like it is nothing
anonymous
  • anonymous
Wolfram seems to agree; the limit doesn't exist. That's my guess
sweetburger
  • sweetburger
LHops international house of limits
KingGeorge
  • KingGeorge
Not only does wolfram not say anything, but Mathematica comes up blank as well.
myininaya
  • myininaya
Does that mean Mathematica doesn't have the programming to find the limit or that the limit does not exist?
myininaya
  • myininaya
i believe the limit does not exist and i also believe we don't need l'hopital (even though it says to use it)
KingGeorge
  • KingGeorge
I think it only tells you that mathematica doesn't have the programming.
anonymous
  • anonymous
Hmm... 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
  • tkhunny
Can 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
  • KingGeorge
This 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
  • tkhunny
Sequence: \((\cos(4n\pi))^{n^2}\;for\; n\in\mathbb{N}\) It's just ones (1s).
anonymous
  • anonymous
I 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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