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]}