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tonykart44

  • 3 years ago

Hi everyone, I'm currently studying for my calculus test tommorow and I have some problems which I have no clue how to solve. 1) Lim (cos 2x)^(csc^2 2x) x-> 0 Any help at all is very much appreciated, as I truly have no idea how to get started this excercise. I'm sorry for not saying this earlier but L'Hopitals rule is not allowed.

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  1. anonymous
    • 3 years ago
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    this looks like \(1^{\infty}\) so you can use l'hopital after you take the log

  2. anonymous
    • 3 years ago
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    when you see an exponential function like this, you usually have to take the log first and work with that one either that, or rewrite as \[e^{\csc^2(2x)\ln(\cos(2x)}\]

  3. anonymous
    • 3 years ago
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    i would take the log, get \[\csc^2(2x)\ln(\cos(2x))\] then rewrite as \[\frac{\ln(\cos(2x))}{\sin^2(2x)}\] to put it in the form \(\frac{0}{0}\) now you can use l'hopital

  4. anonymous
    • 3 years ago
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    when you get your answer to this, make sure to remember to raise \(e\) to that power, as you took the log as the first step

  5. tonykart44
    • 3 years ago
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    Thank you very much for your fast answer but unfortunately we are not allowed to use L'Hoptials rule yet. I'm sorry I didn't specify that, is there another way to solve this problem?

  6. rahul91
    • 3 years ago
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    i dont know whether it help take the whole expression = y and take log on both sides try to simplify

  7. tonykart44
    • 3 years ago
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    That looks like it could work. I'll try that.

  8. tonykart44
    • 3 years ago
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    But isn't that for solving derivatives?

  9. tonykart44
    • 3 years ago
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    Ok so I got this one in the form (1-2sin^2 x)^(1/2sin^2 2x) but i don't know how to go on from there. I know i'm almost there but i don't know how to get rid of the 2x above, I know it's silly, but i just don't see it.

  10. rahul91
    • 3 years ago
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    1-2sin^2x = cos2x

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