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probss Group Title

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

  • 4 months ago
  • 4 months ago

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  1. myininaya Group Title
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    Do you know you can write y as e^ln(y) let me know if that helps brb

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

    • 4 months ago
  3. probss Group Title
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    i think i forgot about it

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

    • 4 months ago
  5. probss Group Title
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    wow, seems it is so complicated.

    • 4 months ago
  6. probss Group Title
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    so, we can just take the value of power of exponential? what say you?

    • 4 months ago
  7. probss Group Title
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    @myininaya

    • 4 months ago
  8. myininaya Group Title
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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

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

    • 4 months ago
  10. myininaya Group Title
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    not 0/dne or dne/inf

    • 4 months ago
  11. zepdrix Group Title
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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?

    • 4 months ago
  12. myininaya Group Title
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    yah

    • 4 months ago
  13. zepdrix Group Title
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    Grr I dunno >.< maybe lego man knows

    • 4 months ago
  14. myininaya Group Title
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    maybe sith knows i seen him around a lot doing math like it is nothing

    • 4 months ago
  15. SithsAndGiggles Group Title
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    Wolfram seems to agree; the limit doesn't exist. That's my guess

    • 4 months ago
  16. sweetburger Group Title
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    LHops international house of limits

    • 4 months ago
  17. KingGeorge Group Title
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    Not only does wolfram not say anything, but Mathematica comes up blank as well.

    • 4 months ago
  18. myininaya Group Title
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    Does that mean Mathematica doesn't have the programming to find the limit or that the limit does not exist?

    • 4 months ago
  19. myininaya Group Title
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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)

    • 4 months ago
  20. KingGeorge Group Title
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    I think it only tells you that mathematica doesn't have the programming.

    • 4 months ago
  21. SithsAndGiggles Group Title
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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.

    • 4 months ago
  22. tkhunny Group Title
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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.

    • 4 months ago
  23. KingGeorge Group Title
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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.

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

    • 4 months ago
  25. DominicNg Group Title
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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.

    • 4 months ago
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