Let #y = pi/2-arctanx#.
Then #coty = cot(pi/2-arctanx) = tan(arctanx) = x# and:
#lim_(x->oo) y(x) = 0#
with #y >0# bacause #arctanx < pi/2#.
So:
#lim_(x->oo) (pi/2-arctanx)^(1/x) = lim_(y->0^+) y^(1/coty)#
As: #1/coty = tany#:
#lim_(x->oo) (pi/2-arctanx)^(1/x) = lim_(y->0^+) y^tany#
Write now the function as:
#y^tany = (e^lny)^tany = e^(lnytany)#
and evaluate the limit:
#lim_(y->0^+) lnytany#
This is in the indeterminate form #oo/0# but we can divide and multiply by #y# to have:
#lim_(y->0^+) (y lny) (tany/y)#
and use the well known limits:
#lim_(y->0) tany/y = 1#
#lim_(y->0^+) ylny = 0#
to have:
#lim_(y->0^+) lnytany = 0#
and then as #e^t# is continuous for every #t in RR#:
#lim_(y->0^+) e^(lnytany) = e^((lim_(y->0^+) lnytany)) =e^0 = 1#
In conclusion:
#lim_(x->oo) (pi/2-arctanx)^(1/x) = 1#
graph{(pi/2-arctanx)^(1/x) [-4.75, 15.25, -4.84, 5.16]}
