1)
Note that the limit is equal to
2)
Hint: use the double-angle identity for
Hi everyone,
I'm currently studying for my calculus test tommorow and I have some problems which I have no clue how to solve. One problem is a limit and the other one is about inverse trig functions.
1) Lim (cos 2x)^(csc^2 2x)
x-> 0
2) 2. Cos(2arccotx)
Any help at all is very much appreciated, as I truly have no idea how to get started on especially the first exercise. Oh and I should mention that we are not allowed to use L'Hopitals rule for these problems.
Thank you very much for your fast answer! But I think I wasn't clear on the first problem, It should be cos(2x) to the power csc^2 2X Sorry if that wasnt clear.
For the second one I did what you said and I got (2X^2/(1+X^2)) -1 does that look alright to you?
No, it's my fault, for some reason I red a fraction in stead of a power.
In case we have to compute . Use the fact that thus we can write the limit as (note ). Because the exponential function is continuous we can write the limit as which can be solved by using l'Hopital's rule.
The solution of the second question is correct. (I assume the 2. was not a factor 2 but just an indication of question 2)
Once again, I will need to annoy you because for these problems, we are actually not allowed to use L'Hôpitals rule. I'm trying to use trig functions to get around that, but I just can't seem to get that right.
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.