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probss

  • one year ago

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

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  1. myininaya
    • one year ago
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    Do you know you can write y as e^ln(y) let me know if that helps brb

  2. myininaya
    • one year ago
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    so what I'm say is you can write \[(\cos(x/2))^{x^2}=e^{\ln((\cos(x/2))^{x^2})}=e^{x^2 \ln(\cos(x/2))}=e^{\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}}\]

  3. probss
    • one year ago
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    i think i forgot about it

  4. myininaya
    • one year ago
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    now you did to find this limit \[\lim_{x \rightarrow \infty}\frac{\ln(\cos(x/2))}{\frac{1}{x^2}}\]

  5. probss
    • one year ago
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    wow, seems it is so complicated.

  6. probss
    • one year ago
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    so, we can just take the value of power of exponential? what say you?

  7. probss
    • one year ago
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    @myininaya

  8. myininaya
    • one year ago
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    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

  9. myininaya
    • one year ago
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    @zepdrix the reason i ask is because i only see it used for cases like inf/inf or 0/0

  10. myininaya
    • one year ago
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    not 0/dne or dne/inf

  11. zepdrix
    • one year ago
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    Oh boy D: Good question.. hmm So we have a problem since ln(cosx) is undefined over and over as x->infinity, yah?

  12. myininaya
    • one year ago
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    yah

  13. zepdrix
    • one year ago
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    Grr I dunno >.< maybe lego man knows

  14. myininaya
    • one year ago
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    maybe sith knows i seen him around a lot doing math like it is nothing

  15. SithsAndGiggles
    • one year ago
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    Wolfram seems to agree; the limit doesn't exist. That's my guess

  16. sweetburger
    • one year ago
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    LHops international house of limits

  17. KingGeorge
    • one year ago
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    Not only does wolfram not say anything, but Mathematica comes up blank as well.

  18. myininaya
    • one year ago
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    Does that mean Mathematica doesn't have the programming to find the limit or that the limit does not exist?

  19. myininaya
    • one year ago
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    i believe the limit does not exist and i also believe we don't need l'hopital (even though it says to use it)

  20. KingGeorge
    • one year ago
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    I think it only tells you that mathematica doesn't have the programming.

  21. SithsAndGiggles
    • one year ago
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    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.

  22. tkhunny
    • one year ago
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    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.

  23. KingGeorge
    • one year ago
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    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.

  24. tkhunny
    • one year ago
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    Sequence: \((\cos(4n\pi))^{n^2}\;for\; n\in\mathbb{N}\) It's just ones (1s).

  25. DominicNg
    • one year ago
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    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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