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## tonykart44 Group Title 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. one year ago one year ago

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1. satellite73 Group Title

this looks like $$1^{\infty}$$ so you can use l'hopital after you take the log

2. satellite73 Group Title

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. satellite73 Group Title

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. satellite73 Group Title

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 Group Title

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 Group Title

i dont know whether it help take the whole expression = y and take log on both sides try to simplify

7. tonykart44 Group Title

That looks like it could work. I'll try that.

8. tonykart44 Group Title

But isn't that for solving derivatives?

9. tonykart44 Group Title

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 Group Title

1-2sin^2x = cos2x