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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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this looks like \(1^{\infty}\) so you can use l'hopital after you take the log
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 \[e^{\csc^2(2x)\ln(\cos(2x)}\]
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

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Other answers:

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
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?
i dont know whether it help take the whole expression = y and take log on both sides try to simplify
That looks like it could work. I'll try that.
But isn't that for solving derivatives?
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.
1-2sin^2x = cos2x

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