- #1

bilco

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## Homework Statement

limit (x->infinity) [pi/2-arctan x]^(1/x)

## Homework Equations

## 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)

which yields:

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.