limit (x->infinity) [pi/2-arctan x]^(1/x)
The Attempt at a Solution
The problem I am attempting to tackle is limit x->infinity [pi/2 - arctan x]^(1/x) ----- (1)
I have attempted an array of solutions, most of which involve very messy derivatives involving L'Hospital's. I've tried taking the natural log of the equation and substituting in k for 1/x etc. After some research and a few hours of attempts the closest to a solution I got was the following:
Using an inverse trig complimentary identity:
arccot x = pi/2 - arctan x, Solving for pi/2 = arccot x + arctan x
Substituting pi/2 for pi/2 into (1):
limit x->infinity [(arccot x + arctan x) - arctan x]^(1/x)
limit x->infinity [arccot x]^(1/x)
from here i used the fact that cot(theta) = x/y. So if x->infinity in arccot x then this imply would mean y->0 and would be 0 on the unit circle.
so we would hav: limit x->infinity [arccot x]^(1/x) = 0^0 which is an indeterminate form.
However we could also consider 0 on the unit circle to be 2pi and then we would have:
limit x->infinity [arccot x]^(1/x) = [2pi]^0 = 1. QED
I feel in this attempt I am cheating the problem by abusing the unit circle, however all other attempts at manipulating 1/x*ln(pi/2-arctanx) didnt yield anything useful.