Question

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.

Answer 1

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

Answer 2

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
\[e^{\csc^2(2x)\ln(\cos(2x)}\]

Answer 3

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

Answer 4

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

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

But isn't that for solving derivatives?

Answer 9

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.

Answer 10

1-2sin^2x = cos2x