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]}