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.

Hey! We 've verified this expert answer for you, click below to unlock the details :)

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.