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

1 Answer
Jun 27, 2018

#lim_(x->oo) (pi/2-arctanx)/x = 0#

Explanation:

Let #y= pi/2-arctanx#. When #x->+oo#, #y->0# and using the trigonometric identity:

#cot(pi/2-theta) = tan theta#

we have:

#cot y = cot(pi/2-arctanx) = tan(arctanx) = x#

Then:

#lim_(x->oo) (pi/2-arctanx)/x = lim_(y->0) y/coty#

#lim_(x->oo) (pi/2-arctanx)/x = lim_(y->0) ytany = 0#

graph{(pi/2-arctanx)/x [-10, 10, -5, 5]}