Find lim_{x to oo} {pi/2 - arctan(x)}^(1/x) ?

1 Answer
Jul 28, 2018

lim_(x->oo) (pi/2-arctanx)^(1/x) = 1

Explanation:

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]}